1 介绍
本博客用来记录灵神力扣题单之二分算法。
2 训练
2.5 最小化最大值
题目50：2513. 最小化两个数组中的最大值
解题思路：二分。仅适合vec1的数，仅适合vec2的数，同时适合vec1和vec2的数。
C++代码如下，
class Solution {
public:
int minimizeSet(int d1, int d2, int cnt1, int cnt2) {
long long d3 = gcd(d1, d2);
d3 = 1ll * d1 * d2 / d3;
function<bool(long long)> check =[&] (long long mid) -> bool {
long long x1 = mid / d2 - mid / d3; //1~mid中仅适合vec1的数
long long x2 = mid / d1 - mid / d3; //1~mid中仅适合vec2的数
long long x3 = mid - mid / d1 - mid / d2 + mid / d3; //1~mid中同时适合vec1和vec2的数
long long a = cnt1, b = cnt2;
a = max(0ll, a - x1);
b = max(0ll, b - x2);
return a + b <= x3;
};
long long left = 0;
long long right = 1e18;
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
import math
class Solution:
def minimizeSet(self, d1: int, d2: int, cnt1: int, cnt2: int) -> int:
def check(mid: int) -> bool:
#1~mid满足上述要求
#适合vec1的数：它不是d1的倍数
x1 = mid - mid // d1
#适合vec2的数：它不是d2的倍数
x2 = mid - mid // d2
#同时适合vec1和vec2的数：它不是d1的倍数并且它不是d2的倍数
overlap = mid // (d1 * d2 // math.gcd(d1, d2))
overlap = int(overlap)
x3 = mid - (mid // d1 + mid // d2 - overlap)
return max(cnt1-x1+x3,0) + max(cnt2-x2+x3,0) <= x3
left = 0
right = int(1e15)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1
else:
left = mid + 1
return res
题目51：LCP 12. 小张刷题计划
解题思路：二分。按照题目编号顺序刷完所有的题目。每天的最大累加值<=mid，需要days天，且days<=m。记录当天的最大用时max_t_in_day。
C++代码如下，
class Solution {
public:
int minTime(vector<int>& time, int m) {
function<bool(int)> check =[&] (int mid) -> bool {
//当天题目用时累加<=mid，需要days天，且days<=m
int curr_time = 0;
int max_t_in_day = 0;
int days = 1;
for (auto t : time) {
curr_time += t;
max_t_in_day = max(max_t_in_day, t);
if (curr_time - max_t_in_day > mid) {
days += 1;
curr_time = t;
max_t_in_day = t;
}
}
return days <= m;
};
int left = 0;
int right = accumulate(time.begin(), time.end(), 0);
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def minTime(self, time: List[int], m: int) -> int:
n = len(time) #不能排序，“按照题目编号顺序”
def check(mid: int) -> bool:
#每天的最大累加值<=mid，需要days天，且days<=m
days = 1
current_time = 0
max_t_in_day = 0
for t in time:
current_time += t
max_t_in_day = max(max_t_in_day, t)
if current_time-max_t_in_day > mid:
days += 1
current_time = t
max_t_in_day = t
return days <= m
left = 0
right = sum(time)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1
else:
left = mid + 1
return res
2.6 最大化最小值
题目52：2517. 礼盒的最大甜蜜度
解题思路：二分。从price中选出差的绝对值>=mid的数，总数为cnt，cnt>=k。
C++代码如下，
class Solution {
public:
int maximumTastiness(vector<int>& price, int k) {
int n = price.size();
sort(price.begin(), price.end());
function<bool(int)> check =[&] (int mid) -> bool {
//从price中选出差的绝对值>=mid的数，总数为cnt，cnt>=k
int cnt = 1;
int prev = price[0];
for (int i = 1; i < n; ++i) {
if (price[i] - prev >= mid) {
cnt += 1;
prev = price[i];
} else {
//pass
}
}
return cnt >= k;
};
int left = 0;
int right = 1e9;
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def maximumTastiness(self, price: List[int], k: int) -> int:
price.sort()
n = len(price)
def check(mid: int) -> bool:
#选出k个数，存在一种选法，这k个数任意两个数的差的绝对值>=mid
#从price中选出差的绝对值>=mid的数，它们的总数cnt>=k
cnt = 1
prev = price[0]
for i in range(1,n):
if price[i] - prev >= mid:
cnt += 1
prev = price[i]
else:
pass
return cnt >= k
left = 0
right = int(1e9)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
left = mid + 1
else:
right = mid - 1
return res
题目53：1552. 两球之间的磁力
解题思路：二分。
C++代码如下，
class Solution {
public:
int maxDistance(vector<int>& position, int m) {
sort(position.begin(), position.end());
int n = position.size();
function<bool(int)> check =[&] (int mid) -> bool {
//从position中选出差的绝对值<=mid的数，它们总数>=m
int cnt = 1;
int prev = position[0];
for (int i = 1; i < n; ++i) {
if (position[i] - prev >= mid) {
cnt += 1;
prev = position[i];
}
}
return cnt >= m;
};
int left = 0;
int right = 1e9;
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def maxDistance(self, position: List[int], m: int) -> int:
position.sort()
n = len(position)
def check(mid: int) -> bool:
#从position中选出差的绝对值>=mid的数，它们的总数cnt>=m
cnt = 1
prev = position[0]
for i in range(1,n):
if position[i] - prev >= mid:
cnt += 1
prev = position[i]
return cnt >= m
left = 0
right = int(1e9)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
left = mid + 1
else:
right = mid - 1
return res
题目54：2812. 找出最安全路径
解题思路：二分+bfs。
C++代码如下，
class Solution {
public:
int maximumSafenessFactor(vector<vector<int>>& grid) {
int n = grid.size();
int m = grid[0].size();
int dirs[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
//先做一遍bfs，d[i][j]表示到小偷的最短距离
vector<vector<int>> d(n, vector<int>(m, -1));
queue<pair<int,int>> q;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == 1) {
q.push(make_pair(i,j));
d[i][j] = 0;
}
}
}
while (!q.empty()) {
auto t = q.front();
q.pop();
int x = t.first, y = t.second;
for (int k = 0; k < 4; ++k) {
int nx = x + dirs[k][0];
int ny = y + dirs[k][1];
if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
if (d[nx][ny] != -1) continue;
d[nx][ny] = d[x][y] + 1;
q.push(make_pair(nx,ny));
}
}
function<bool(int)> check =[&] (int mid) -> bool {
//只能走d[i][j]>=mid的格子，能到达终点
if (d[0][0] < mid) return false; //特判
vector<vector<bool>> st(n, vector<bool>(m, false));
queue<pair<int,int>> q;
q.push(make_pair(0,0));
st[0][0] = true;
while (!q.empty()) {
auto t = q.front();
q.pop();
int x = t.first, y = t.second;
if (x == n-1 && y == m-1) { //走到了终点
return true;
}
for (int k = 0; k < 4; ++k) {
int nx = x + dirs[k][0];
int ny = y + dirs[k][1];
if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
if (d[nx][ny] < mid) continue;
if (st[nx][ny]) continue;
q.push(make_pair(nx,ny));
st[nx][ny] = true;
}
}
return false;
};
int left = 0;
int right = n + m;
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
n = len(grid)
m = len(grid[0])
#先做一遍bfs，d[i][j]表示到小偷的最小距离
d = [[-1] * m for _ in range(n)]
q = collections.deque()
for i in range(n):
for j in range(m):
if grid[i][j] == 1:
q.append([i,j])
d[i][j] = 0
dirs = [[-1,0],[1,0],[0,-1],[0,1]]
while len(q) > 0:
x, y = q.popleft()
for k in range(4):
nx = x + dirs[k][0]
ny = y + dirs[k][1]
if nx < 0 or nx >= n or ny < 0 or ny >= m:
continue
if d[nx][ny] != -1:
continue
d[nx][ny] = d[x][y] + 1
q.append([nx,ny])
def check(mid: int) -> bool:
#只能走d[i][j]>=mid的格子
if d[0][0] < mid: #特判
return False
q = collections.deque()
st = [[False]*m for _ in range(n)]
q.append([0,0])
st[0][0] = True
while len(q) > 0:
x, y = q.popleft()
if x == n-1 and y == m-1:
return True #走到了终点
for k in range(4):
nx = x + dirs[k][0]
ny = y + dirs[k][1]
if nx < 0 or nx >= n or ny < 0 or ny >= m:
continue
if d[nx][ny] < mid:
continue
if st[nx][ny]:
continue
q.append([nx,ny])
st[nx][ny] = True
return False
left = 0
right = n + m
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
left = mid + 1
else:
right = mid - 1
return res
题目55：2528. 最大化城市的最小电量
解题思路：二分+贪心+差分+前缀和。当nums[i]，不满足条件，在i+r处建立供电站。
C++代码如下，
class Solution {
public:
long long maxPower(vector<int>& stations, int r, int k) {
int n = stations.size();
vector<long long> s(n+1,0);
for (int i = 1; i < n+1; ++i) {
s[i] = s[i-1] + stations[i-1];
}
vector<long long> nums(n,0);
for (int i = 0; i < n; ++i) {
int end = min(i+r+1, n);
int start = max(i-r, 0);
nums[i] = s[end] - s[start];
}
function<bool(long long)> check =[&] (long long mid) -> bool {
//每个城市的电量都大于等于mid，需要建造cnt个供电站，最终cnt <= k
vector<long long> diff(n, 0);
long long sumd = 0;
long long cnt = 0;
for (int i = 0; i < n; ++i) {
sumd += diff[i];
long long x = mid - sumd - nums[i];
if (x > 0) {
cnt += x;
sumd += x;
if (i+r*2+1 < n) {
diff[i+r*2+1] -= x;
}
}
}
return cnt <= k;
};
long long left = ranges::min(nums);
long long right = left + k + 1;
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (check(mid)) {
res = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def maxPower(self, stations: List[int], r: int, k: int) -> int:
n = len(stations)
s = [0] * (n+1)
for i in range(1,n+1):
s[i] = s[i-1] + stations[i-1]
nums = [0] * n
for i in range(n):
nums[i] = s[min(i+r+1,n)]-s[max(i-r,0)] #[i-r,i+r] ===> (i-r,i+r+1]
def check(mid):
#使得每个城市的电量都大于等于mid，需要新建cnt个供电站
diff = [0] * n
sumd = 0 #差分数目累加和
cnt = 0 #需要新建的供电站数目
for i,num in enumerate(nums):
sumd += diff[i]
y = mid - sumd - num
if y > 0:
cnt += y #需要新建y座供电站
if cnt > k:
return False
sumd += y
if i+r*2+1 < n:
diff[i+r*2+1] -= y
return True
left = min(nums)
right = left + k + 1
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
left = mid + 1
else:
right = mid - 1
return res
2.7 第K小/大
题目56：378. 有序矩阵中第 K 小的元素
解题思路：二分。依据lower_bound()，计算大于等于mid的最小值。left = matrix[0][0]，right = matrix[n-1][n-1]，从最后一行第一列开始遍历。
C++代码如下，
class Solution {
public:
int kthSmallest(vector<vector<int>>& matrix, int k) {
int n = matrix.size();
function<bool(int)> check =[&] (int mid) -> bool {
//至少要找到k个小于等于mid的元素
int i = n-1, j = 0;
int count = 0; //矩阵中有多少个数小于等于mid
while (i >= 0 && j < n) {
if (matrix[i][j] <= mid) {
count += i+1;
j += 1;
} else {
i -= 1;
}
}
return count >= k;
};
int left = matrix[0][0];
int right = matrix[n-1][n-1];
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid; //求满足条件的最小值
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
n = len(matrix)
def check(mid: int) -> int:
#矩阵中小于等于mid的数的个数，它大于等于k
i = n-1
j = 0
count = 0
#从最后一行第一列开始，计算每一列中小于等于mid的个数
while i >= 0 and j < n:
if matrix[i][j] <= mid:
count += i + 1
j += 1
else:
i -= 1
return count >= k
left = matrix[0][0]
right = matrix[n-1][n-1]
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid #计算大于等于k的最小值
right = mid - 1
else:
left = mid + 1
return res
题目57：668. 乘法表中第k小的数
解题思路：二分。统计矩阵中<=mid的数目cnt，最终cnt >= k。求mid的最小值。
C++代码如下，
class Solution {
public:
int findKthNumber(int n, int m, int k) {
function<bool(int)> check =[&] (int mid) -> bool {
int cnt = 0;
int i = n, j = 1;
while (i >= 1 && j <= m) {
if (i*j <= mid) {
cnt += i;
j += 1;
} else {
i -= 1;
}
}
return cnt >= k;
};
int left = 1;
int right = n * m;
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid-1;
} else {
left = mid+1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def findKthNumber(self, n: int, m: int, k: int) -> int:
def check(mid: int) -> bool:
#至少有k个数不超过mid
cnt = 0
i = n
j = 1
while i >= 1 and j <= m:
if i*j <= mid:
cnt += i
j += 1
else:
i -= 1
return cnt >= k
left = 1
right = n * m
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid #至少有k个数不超过mid
#求mid的最小值
right = mid - 1
else:
left = mid + 1
return res
题目58：719. 找出第 K 小的数对距离
解题思路：二分+双指针。至少k个数对<=mid。求最小值。
C++代码如下，
class Solution {
public:
int smallestDistancePair(vector<int>& nums, int k) {
int n = nums.size();
sort(nums.begin(), nums.end());
function<bool(int)> check =[&] (int mid) -> bool {
int cnt = 0;
int j = 0;
for (int i = 0; i < n; ++i) {
while (j < n && nums[j]-nums[i] <= mid) {
j += 1;
}
cnt += j - i - 1;
}
return cnt >= k;
};
int left = 0;
int right = ranges::max(nums) - ranges::min(nums);
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) { //至少k个数对<=mid
res = mid;
right = mid - 1; //求最小值
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def smallestDistancePair(self, nums: List[int], k: int) -> int:
nums.sort()
n = len(nums)
def check(mid: int) -> bool:
j = 0
cnt = 0
for i in range(n):
while j < n and nums[j] - nums[i] <= mid:
j += 1
cnt += j - i - 1
return cnt >= k
left = 0
right = max(nums) - min(nums)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid): #至少k个数对<=mid
res = mid
right = mid - 1 #求最小值
else:
left = mid + 1
return res
题目59：878. 第 N 个神奇数字
解题思路：二分+容斥原理。
C++代码如下，
class Solution {
public:
int nthMagicalNumber(int n, int a, int b) {
int c = a * b / gcd(a, b);
function<bool(long long)> check =[&] (long long mid) -> bool {
long long cnt = mid / a + mid / b - mid / c;
return cnt >= n;
};
long long left = n;
long long right = 1e14;
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (check(mid)) { //至少k个数<=mid
res = mid;
right = mid - 1;//求最小值
} else {
left = mid + 1;
}
}
return res % (int)(1e9+7);
}
};
python3代码如下，
import math
class Solution:
def nthMagicalNumber(self, n: int, a: int, b: int) -> int:
c = a * b // math.gcd(a,b) #a和b的最小公倍数
def check(mid: int) -> bool:
cnt = mid // a + mid // b - mid // c #容斥原理
return cnt >= n
left = 0
right = int(1e14)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid): #求至少n个数满足要求
res = mid
right = mid - 1 #求最小值
else:
left = mid + 1
return res % int(1e9 + 7)
题目60：1201. 丑数 III
解题思路：二分+容斥定理。
C++代码如下，
class Solution {
public:
long long lcm(long long x, long long y) {
return x * y / gcd(x, y);
}
int nthUglyNumber(int n, int a, int b, int c) {
long long x = lcm(a, b);
long long y = lcm(a, c);
long long z = lcm(b, c);
long long u = lcm(x, c);
function<bool(int)> check =[&] (long long mid) -> bool {
//至少有n个数
long long cnt = mid / a + mid / b + mid / c - mid / x - mid / y - mid / z + mid / u;
return cnt >= n;
};
int left = n;
int right = 2e9 + 10;
int res = -1;
while (left <= right) {
long long mid = (1ll * left + right) / 2;
if (check(mid)) { //至少有n个数
res = mid;
right = mid - 1; //求最小值
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
import math
class Solution:
def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
def lcm(x: int, y: int) -> int:
return x * y // math.gcd(x,y)
x = lcm(a, b)
y = lcm(a, c)
z = lcm(b, c)
u = lcm(x, c)
def check(mid: int) -> bool:
#print(f"mid={mid}.")
cnt = mid // a + mid // b + mid // c - mid // x - mid // y - mid // z + mid // u
#print(f"cnt={cnt},cnt>=n ==> {cnt>=n}.")
return cnt >= n
left = n
right = int(2e9 + 10)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid): #至少n个数满足要求
res = mid
right = mid - 1 #求最小值
else:
left = mid + 1
return res
题目61：793. 阶乘函数后 K 个零
解题思路：二分+数论。
（1）x!中末尾0的个数：它是由因子10的个数决定的。因为10是由2和5相乘而来的，而阶乘中2的因子总是比5多。因此，末尾0的个数等于因子5的出现次数。
（2）为了计算阶乘中包含多少个5的因子，我们可以从5的倍数入手。每当一个数是5的倍数时，它会为阶乘贡献至少一个5的因子。接着。如果一个数是25的倍数，它会多贡献一个5的因子。类似地，125的倍数会再贡献一个5的因子。故x!中5的因子的个数为，x//5 + x//25 + x//125 + ...。
分别表示：
- 1~x中所有能被5整除的数的个数。
- 1~x中所有能被35整除的数的个数。
- 1~x中所有能被125整除的数的个数。
- 以此类推……
C++代码如下，
class Solution {
public:
long long f(long long x) {
long long res = 0;
while (x > 0) {
res += x / 5;
x /= 5;
}
return res;
}
int search_left(int k) {
//f[x]=k，求最小的x
long long left = 0;
long long right = 5ll * (k + 1);
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (f(mid) >= k) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
int preimageSizeFZF(int k) {
return search_left(k+1) - search_left(k);
}
};
python3代码如下，
class Solution:
def preimageSizeFZF(self, k: int) -> int:
#x!末尾中0的数量，等价于因子5的数目
#f[x]=x//5+x//25+x//125+...
def f(x: int) -> int:
#计算f[x]
res = 0
while x > 0:
res += x // 5
x //= 5
return res
def search_left(k: int) -> int:
#f[x]>=k，求x的最小值
left = 0
right = 5 * (k+1)
res = -1
while left <= right:
mid = (left + right) // 2
if f(mid) >= k:
res = mid
right = mid - 1
else:
left = mid + 1
return res
def search_right(k: int) -> int:
#f[x] > k, 求x的最小值
left = 0
right = 5 * (k + 1)
res = -1
while left <= right:
mid = (left + right) // 2
if f(mid) > k:
res = mid
right = mid - 1
else:
left = mid + 1
return res
return search_right(k) - search_left(k)
题目62：373. 查找和最小的 K 对数字
解题思路：贪心。
（1）堆的初始化：将nums1中的前k个元素与nums2[0]配对，放入一个最小堆中。堆中的每个元素是一个三元组(total_sum, i, j)，表示nums1[i] + nums2[j]的和、索引i和j。
（2）从堆中取数对：每次从堆中取出和最小的数对(nums1[i], nums2[j])，然后将下一个数对(nums1[i], nums2[j+1])放入堆中（如果j+1是有效的索引）。
C++代码如下，
class Solution {
public:
vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
struct node {
int s;
int i;
int j;
bool operator> (const node& node1) const {
return s > node1.s;
}
node(int s, int i, int j) : s(s), i(i), j(j) {}
};
int n = nums1.size();
int m = nums2.size();
priority_queue<node,vector<node>,greater<node>> hp;
for (int i = 0; i < min(n, k); ++i) {
node t(nums1[i]+nums2[0],i,0);
hp.push(t);
}
vector<vector<int>> res;
while (k > 0 && !hp.empty()) {
auto t = hp.top();
hp.pop();
int total_sum = t.s;
int i = t.i;
int j = t.j;
res.push_back({nums1[i],nums2[j]});
k -= 1;
if (j+1 < m) {
hp.push(node(nums1[i]+nums2[j+1],i,j+1));
}
}
return res;
}
};
python3代码如下，
import heapq
class Solution:
def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
hp = []
for i in range(min(k,len(nums1))):
heapq.heappush(hp, (nums1[i]+nums2[0], i, 0))
res = []
while k > 0 and len(hp) > 0:
total_sum, i, j = heapq.heappop(hp)
res.append([nums1[i], nums2[j]])
k -= 1
if j + 1 < len(nums2):
heapq.heappush(hp, (nums1[i]+nums2[j+1], i, j+1))
return res
解题思路：二分+bfs+小根堆。
C++代码如下，
class Solution {
public:
int kthSmallest(vector<vector<int>>& mat, int k) {
int n = mat.size();
int m = mat[0].size();
struct node {
int s;
vector<int> idxes;
node(int in_s, vector<int> in_idxes) : s(in_s), idxes(in_idxes) {}
bool operator> (const node& node1) const { //小根堆，它重载大于号
return s > node1.s;
}
};
function<bool(int)> check =[&] (int mid) -> bool {
priority_queue<node,vector<node>,greater<node>> hp;
int s = 0;
for (auto& row : mat) {
s += row[0];
}
hp.push(node(s,vector<int>(n,0)));
set<vector<int>> seen;
int cnt = 0;
while (!hp.empty()) {
node t = hp.top();
hp.pop();
int curr_s = t.s;
vector<int> curr_idxes = t.idxes;
cnt += 1;
if (curr_s > mid) {
break;
}
if (cnt >= k) { //提前判断
return true;
}
for (int i = 0; i < n; ++i) {
int j = curr_idxes[i];
if (j + 1 < m) {
vector<int> new_idxes = curr_idxes;
new_idxes[i] += 1;
int new_s = curr_s - mat[i][j] + mat[i][j+1];
if (seen.find(new_idxes) == seen.end()) {
seen.insert(new_idxes);
hp.push(node(new_s,new_idxes));
}
}
}
}
return false;
};
int left = 0;
int right = 0;
for (auto& row : mat) {
left += row[0];
right += row.back();
}
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1; //求最小值
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
import heapq
class Solution:
def kthSmallest(self, mat: List[List[int]], k: int) -> int:
n, m = len(mat), len(mat[0])
def check(mid: int) -> bool:
#数组和小于等于mid，这样的数组有cnt个
hp = [(sum([row[0] for row in mat]), [0]*n)]
visited = set(tuple([0]*n))
#heapq.heapify(hp)
cnt = 0 #
while len(hp) > 0:
curr_s, curr_idxes = heapq.heappop(hp)
if curr_s > mid:
break
cnt += 1
if cnt >= k: #提前判断
return True
for i in range(n):
j = curr_idxes[i]
if j + 1 < m:
next_idxes = list(curr_idxes)
next_idxes[i] += 1
next_s = curr_s - mat[i][j] + mat[i][j+1]
next_idxes = tuple(next_idxes)
if next_idxes not in visited:
heapq.heappush(hp, (next_s, list(next_idxes)))
visited.add(next_idxes)
return False
left = sum([row[0] for row in mat])
right = sum([row[-1] for row in mat])
#print(f"left = {left}, right = {right}.")
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid): #和小于等于mid的数组个数，它大于等于k
res = mid
right = mid - 1 #求最小值
else:
left = mid + 1
return res
题目64：786. 第 K 个最小的质数分数
解题思路：直接模拟。
C++代码如下，
class Solution {
public:
vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {
struct node {
double val;
int a;
int b;
node(double in_val, int in_a, int in_b) : val(in_val), a(in_a), b(in_b) {}
bool operator< (const node& node1) const {
return val < node1.val;
}
};
int n = arr.size();
vector<node> new_arr;
for (int i = 0; i < n; ++i) {
for (int j = i+1; j < n; ++j) {
double val = 1.0 * arr[i] / arr[j];
new_arr.push_back(node(val,arr[i],arr[j]));
}
}
sort(new_arr.begin(), new_arr.end());
int x = new_arr[k-1].a;
int y = new_arr[k-1].b;
return {x,y};
}
};
python3代码如下，
class Solution:
def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
n = len(arr)
new_arr = []
for i in range(n):
for j in range(i+1,n):
new_arr.append([arr[i]/arr[j],arr[i],arr[j]])
new_arr.sort()
x, y = new_arr[k-1][1], new_arr[k-1][2]
return [x,y]
题目65：3116. 单面值组合的第 K 小金额
解题思路：二分+容斥原理。
假设有n个集合A1, A2, ..., An，它们的并集用符号∪表示，那么它们的并集的元素个数可以用以下公式计算：
|A1 ∪ A2 ∪ ... ∪ An| =
|A1| + |A2| + ... + |An| // 单个集合的元素个数之和
- |A1 ∩ A2| - |A1 ∩ A3| - ... // 两两相交部分的元素个数之和
+ |A1 ∩ A2 ∩ A3| + ... // 三三相交部分的元素个数之和 - ...
+ (-1)^(n-1) * |A1 ∩ A2 ∩ ... ∩ An| // 所有集合相交部分的元素个数
C++代码如下，
class Solution {
public:
long long findKthSmallest(vector<int>& coins, int k) {
int n = coins.size();
long long m = 1ll << n;
function<bool(long long)> check =[&](long long mid) {
//1~mid符合要求的数的个数cnt>=k
long long cnt = 0;
for (int i = 1; i < m; ++i) {
bool flag = true;
int bit_1_count = 0;
long long curr = 1; //当前选法的最小公倍数
for (int j = 0; j < n; ++j) {
if (i >> j & 1) {
curr = lcm(curr, 1ll * coins[j]);
bit_1_count += 1;
if (curr > mid) {
flag = false;
break;
}
}
}
if (flag) {
if (bit_1_count % 2) {
cnt += mid / curr;
} else {
cnt -= mid / curr;
}
}
}
return cnt >= k;
};
long long left = 0;
long long right = 1ll * k * ranges::min(coins);
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1; //求最小值
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def findKthSmallest(self, coins: List[int], k: int) -> int:
n = len(coins)
def check(target: int):
#容斥原理
cnt = 0
for i in range(1,1<<n): #至少选择1个数
curr = 1 #当前选法的最小公倍数
for j in range(n):
if i >> j & 1:
curr = math.lcm(curr, coins[j])
if curr > mid:
break
else:
if i.bit_count() % 2 == 1:
cnt += mid // curr
else:
cnt -= mid // curr
return cnt >= k
left = 0
right = min(coins) * k
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1
else:
left = mid + 1
return res
题目66：3134. 找出唯一性数组的中位数
解题思路：二分+滑动窗口。第k小的数，k从1开始编号。
C++代码如下，
class Solution {
public:
int medianOfUniquenessArray(vector<int>& nums) {
int n = nums.size();
long long m = (1ll + n) * n / 2;
long long k = (m + 1) / 2;
function<bool(int)> check =[&] (int mid) -> bool {
//区间内不同的数小于等于mid，这样的区间个数大于等于k
unordered_map<int,int> map_val_cnt;
map_val_cnt[nums[0]] += 1;
int curr = 1; //当前区间内不同的数的个数
long long res = 0;
int i = 0;
int j = 0;
while (i < n) {
while (j < n && curr <= mid) {
j += 1;
if (j < n) {
map_val_cnt[nums[j]] += 1;
if (map_val_cnt[nums[j]] == 1) {
curr += 1;
}
}
}
res += j - i;
if (res >= k) { //提前退出
return true;
}
map_val_cnt[nums[i]] -= 1;
if (map_val_cnt[nums[i]] == 0) {
curr -= 1;
}
i += 1;
}
return false;
};
int left = 0;
int right = n;
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1;//求最小值
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def medianOfUniquenessArray(self, nums: List[int]) -> int:
n = len(nums)
n = (1 + n) * n // 2 #所有数组的总数
k = (n + 1) // 2 #第k个数，从1开始编号
#print(f"k = {k}.")
def check(mid: int) -> bool:
#区间内不同数的个数小于等于mid，这样的区间有cnt个，cnt>=k
#print(f"mid = {mid}.")
cnt = 0
n = len(nums)
i = 0
j = 0
map_val_cnt = collections.defaultdict(int)
curr = 1 #当前滑动区间内，不相同的数的数目
map_val_cnt[nums[0]] += 1
while i < n:
while j < n and curr <= mid:
j += 1
if j < n:
map_val_cnt[nums[j]] += 1
if map_val_cnt[nums[j]] == 1:
curr += 1
cnt += j - i #[i,j)
#print(f"i = {i}, j = {j}.")
map_val_cnt[nums[i]] -= 1
if map_val_cnt[nums[i]] == 0:
curr -= 1
i += 1
#print(f"cnt = {cnt}.")
return cnt >= k
left = 1
right = len(nums)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1 #求最小值
else:
left = mid + 1
return res
题目67：2040. 两个有序数组的第 K 小乘积
解题思路：分负数、正数部分，然后再进行二分。
C++代码如下，
class Solution {
public:
long long func1(vector<int>& pos1, vector<int>& pos2, vector<int>& neg1, vector<int>& neg2, long long k) {
function<bool(long long)> check =[&] (long long mid) -> bool {
//乘积小于等于mid的数对，数目为cnt，cnt>=k
long long cnt = 0;
for (int x : pos1) {
long long y = mid / x;
// if (y > INT_MAX) {
// continue;
// }
auto iter = upper_bound(neg2.begin(), neg2.end(), y);
cnt += distance(neg2.begin(), iter);
if (cnt >= k) {
return true; //提前判断
}
}
for (int x : neg1) {
long long y = mid / (-x);
// if (y > INT_MAX) {
// continue;
// }
auto iter = upper_bound(pos2.begin(), pos2.end(), y);
cnt += distance(pos2.begin(), iter);
if (cnt >= k) {
return true; //提前判断
}
}
return false;
};
long long left = 0;
long long right = 1e10;
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
long long func2(vector<int>& pos1, vector<int>& pos2, vector<int>& neg1, vector<int>& neg2, long long k) {
function<bool(long long)> check =[&] (long long mid) -> bool {
//乘积小于等于mid的数对，数目为cnt，cnt>=k
long long cnt = 0;
for (int x : pos1) {
long long y = mid / x;
// if (y > INT_MAX) {
// continue;
// }
auto iter = upper_bound(pos2.begin(), pos2.end(), y);
cnt += distance(pos2.begin(), iter);
if (cnt >= k) {
return true; //提前判断
}
}
for (int x : neg1) {
long long y = mid / (-x);
// if (y > INT_MAX) {
// continue;
// }
auto iter = upper_bound(neg2.begin(), neg2.end(), y);
cnt += distance(neg2.begin(), iter);
if (cnt >= k) {
return true; //提前判断
}
}
return false;
};
long long left = 0;
long long right = 1e10;
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
long long kthSmallestProduct(vector<int>& nums1, vector<int>& nums2, long long k) {
int n = nums1.size(), m = nums2.size();
auto iter = lower_bound(nums1.begin(), nums1.end(), 0);
int i0 = distance(nums1.begin(), iter); //nums1中负数的个数
iter = upper_bound(nums1.begin(), nums1.end(), 0);
int i1 = distance(nums1.begin(), iter); //nums1中非正数的个数
iter = lower_bound(nums2.begin(), nums2.end(), 0);
int j0 = distance(nums2.begin(), iter); //nums2中负数的个数
iter = upper_bound(nums2.begin(), nums2.end(), 0);
int j1 = distance(nums2.begin(), iter); //nums2中非正数的个数
vector<int> neg1, neg2, pos1, pos2;
neg1.insert(neg1.end(), nums1.begin(), nums1.begin()+i0);
pos1.insert(pos1.end(), nums1.begin()+i1, nums1.end());
neg2.insert(neg2.end(), nums2.begin(), nums2.begin()+j0);
pos2.insert(pos2.end(), nums2.begin()+j1, nums2.end());
// cout << "neg1: ";
// for (auto x : neg1) cout << x << " ";
// cout << endl;
// cout << "pos1: ";
// for (auto x : pos1) cout << x << " ";
// cout << endl;
// cout << "neg2: ";
// for (auto x : neg2) cout << x << " ";
// cout << endl;
// cout << "pos2: ";
// for (auto x : pos2) cout << x << " ";
// cout << endl;
//将neg2符号变成正的，方便search
for (int& x : neg2) {
x = -x;
}
reverse(neg2.begin(), neg2.end());
long long cnt1 = 1ll * i0 * (m -j1) + 1ll * j0 * (n-i1); //乘积为负数的个数
long long cnt2 = 1ll * i0 * j0 + 1ll * (n-i1) * (m-j1); //乘积为正数的个数
long long cnt3 = 1ll * n * m - cnt1 - cnt2; //乘积为0的个数
//cout << "cnt1 = " << cnt1 << ", cnt2 = " << cnt2 << ", cnt3 = " << cnt3 << endl;
if (k <= cnt1) {
k = cnt1 - k + 1;
//在负数中找到第k小的数
return -func1(pos1, pos2, neg1, neg2, k);
} else if (k <= cnt1 + cnt3) {
return 0;
} else {
k -= cnt1 + cnt3;
//在正数中找到第k小的数
return func2(pos1, pos2, neg1, neg2, k);
}
}
};
python3代码如下，
#self
class Solution:
def func1(self, pos1, pos2, neg1, neg2, k):
def check(mid: int) -> bool:
#乘积小于等于mid的数对，个数为cnt，cnt>=k
cnt = 0
for x in pos1:
y = bisect.bisect_right(neg2, mid // x)
cnt += y
for x in neg1:
x = -x
y = bisect.bisect_right(pos2, mid // x)
cnt += y
return cnt >= k
left = 0
right = int(1e10)
res = inf
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1
else:
left = mid + 1
return res
def func2(self, pos1, pos2, neg1, neg2, k):
def check(mid: int) -> bool:
#乘积小于等于mid的数对，它的个数为cnt，cnt>=k
cnt = 0
for x in pos1:
y = bisect.bisect_right(pos2, mid // x)
cnt += y
for x in neg1:
x = -x
y = bisect.bisect_right(neg2, mid // x)
cnt += y
return cnt >= k
left = 0
right = int(1e10)
res = inf
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1
else:
left = mid + 1
return res
def kthSmallestProduct(self, nums1: List[int], nums2: List[int], K: int) -> int:
#求num1*num2中第k小的数
i0 = bisect.bisect_left(nums1, 0) #nums1中的负数的个数
i1 = bisect.bisect_right(nums1, 0) #nums1中非正数的个数
j0 = bisect.bisect_left(nums2, 0) #nums2中的负数的个数
j1 = bisect.bisect_right(nums2, 0) #nums2中非正数的个数
n, m = len(nums1), len(nums2)
cnt1 = i0 * (m-j1) + j0 * (n-i1) #负数乘积的个数
cnt2 = i0 * j0 + (n-i1) * (m-j1) #正数乘积的个数
cnt3 = n * m - cnt1 - cnt2 #乘积为0的个数
neg1, neg2 = nums1[0:i0], nums2[0:j0]
pos1, pos2 = nums1[i1:], nums2[j1:]
#将neg2变成正数方便search
for i in range(len(neg2)):
neg2[i] = -neg2[i]
neg2.reverse()
#print(f"cnt1 = {cnt1}, cnt2 = {cnt2}, cnt3 = {cnt3}.")
k = K
if k <= cnt1:
k = cnt1 - k + 1
#求乘积为负数中的第k小的数
return -self.func1(pos1, pos2, neg1, neg2, k)
elif k <= cnt1 + cnt3:
return 0
else:
k -= cnt1 + cnt3
#求乘积为整数中第k小的数
return self.func2(pos1, pos2, neg1, neg2, k)
#参考版本
class Solution:
def kthSmallestProduct(self, nums1: List[int], nums2: List[int], K: int) -> int:
def count_pairs(neg1, neg2, pos1, pos2, t, k, is_neg):
#乘积<=t的个数
cnt = 0
if is_neg:
for v in pos1:
cnt += bisect_right(neg2, t // v) #nge2是一个仅包含正数的数组
for v in neg1:
cnt += bisect_right(pos2, t // -v)
else:
for v in pos1:
cnt += bisect_right(pos2, t // v)
for v in neg1:
cnt += bisect_right(neg2, t // -v)
return cnt >= k
def binary_search_neg(neg1, neg2, pos1, pos2, k):
def condition(t):
return count_pairs(neg1, neg2, pos1, pos2, t, k, True)
low, high = 0, 10**10
while low < high:
mid = (low + high) // 2
if condition(mid):
high = mid
else:
low = mid + 1
return low
def binary_search_pos(neg1, neg2, pos1, pos2, k):
def condition(t):
return count_pairs(neg1, neg2, pos1, pos2, t, k, False)
low, high = 0, 10**10
while low < high:
mid = (low + high) // 2
if condition(mid):
high = mid
else:
low = mid + 1
return low
#start from here
n1, n2 = len(nums1), len(nums2)
p10 = bisect_left(nums1, 0) #nums1中有p10个负数
p11 = bisect_left(nums1, 1) #nums1中有n1-p11个正数
p20 = bisect_left(nums2, 0) #nums2中有p20个负数
p21 = bisect_left(nums2, 1) #nums2中有n2-p21个正数
neg = p10 * (n2 - p21) + p20 * (n1 - p11) # 负数乘积个数
pos = p10 * p20 + (n1 - p11) * (n2 - p21) # 正数乘积个数
zero = n1 * n2 - neg - pos # 乘积为 0 的个数
pos1, pos2 = nums1[p11:], nums2[p21:]
neg1, neg2 = nums1[:p10], nums2[:p20]
# 将 neg2 中的元素取负值
neg2 = [-x for x in neg2]
neg2.reverse()
k = K
if k <= neg:
k = neg - k + 1 #为什么是neg-(k-1)
#总共neg个数，分成两部分k个数和neg-k个数。neg-k+1个数。
return -binary_search_neg(neg1, neg2, pos1, pos2, k)
if k <= neg + zero:
return 0
k -= neg + zero
return binary_search_pos(neg1, neg2, pos1, pos2, k)
题目68：2386. 找出数组的第 K 大和
解题思路：思维转换+二分。
C++代码如下，
class Solution {
public:
long long kSum(vector<int>& nums, int k) {
//最大的子序列的和
long long s = 0;
for (int i = 0; i < nums.size(); ++i) {
if (nums[i] > 0) {
s += nums[i];
} else {
nums[i] = -nums[i];
}
}
sort(nums.begin(), nums.end()); //排序nums数组
///求数组nums中第k小的子序列的和
function<bool(long long, long long)> check =[&](long long mid, long long k) -> bool {
long long cnt = 1;
function<void(int,long long)> dfs =[&] (int i, long long curr) -> void {
if (cnt == k || i == nums.size() || curr + nums[i] > mid) {
return;
}
cnt += 1;
dfs(i+1, curr+nums[i]);
dfs(i+1, curr);
return;
};
dfs(0,0);
return cnt == k;
};
long long left = 0;
long long right = 0;
for (auto x : nums) right += x;
long long res = -1;
while (left <= right) {
long long mid = (left + right) / 2;
if (check(mid, k)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return s - res;
}
};
python3代码如下，
class Solution:
def kSum(self, nums: List[int], k: int) -> int:
#最大的数组和
s = 0
for i in range(len(nums)):
if nums[i] > 0:
s += nums[i]
else:
nums[i] = -nums[i]
nums.sort() #排序nums数组
###在nums中找到第k小的子序列和
def check(mid: int, k: int) -> bool:
#从nums中找到小于等于mid的子序列和，这样的子序列个数>=k
cnt = 1
def dfs(i: int, curr: int) -> None:
nonlocal cnt
if cnt == k or i == len(nums) or curr + nums[i] > mid:
return
cnt += 1
dfs(i+1, curr+nums[i])
dfs(i+1, curr)
return
dfs(0,0)
return cnt == k
left = 0
right = sum(nums)
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid, k):
res = mid
right = mid - 1
else:
left = mid + 1
return s - res
题目69：1508. 子数组和排序后的区间和
解题思路：简单题简单做。
C++代码如下，
class Solution {
public:
int rangeSum(vector<int>& nums, int n, int left, int right) {
const int mod = 1e9 + 7;
vector<int> res;
for (int i = 0; i < n; ++i) {
int curr = 0;
for (int j = i; j < n; ++j) {
curr += nums[j];
res.push_back(curr);
}
}
sort(res.begin(), res.end());
int ans = 0;
for (int i = left-1; i < right; ++i) {
ans += res[i] % mod;
ans %= mod;
}
return ans;
}
};
python3代码如下，
class Solution:
def rangeSum(self, nums: List[int], n: int, left: int, right: int) -> int:
#子数组的和
res = []
n = len(nums)
for i in range(n):
curr = 0
for j in range(i,n):
curr += nums[j]
res.append(curr)
res.sort()
ans = sum(res[left-1:right])
ans %= int(1e9+7)
return ans
2.8 其它
题目70：2476. 二叉搜索树最近节点查询
解题思路：二分。
C++代码如下，
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> closestNodes(TreeNode* root, vector<int>& queries) {
vector<int> nums;
function<void(TreeNode*)> dfs = [&] (TreeNode* node) -> void {
if (node == nullptr) {
return;
}
dfs(node->left);
nums.push_back(node->val);
dfs(node->right);
return;
};
dfs(root);
vector<vector<int>> res;
for (auto query : queries) {
vector<int> t = {-1,-1};
auto iter = lower_bound(nums.begin(), nums.end(), query+1);
int idx1 = distance(nums.begin(), iter);
idx1 -= 1;
if (0 <= idx1 && idx1 < nums.size()) {
t[0] = nums[idx1];
}
iter = lower_bound(nums.begin(), nums.end(), query);
int idx2 = distance(nums.begin(), iter);
if (0 <= idx2 && idx2 < nums.size()) {
t[1] = nums[idx2];
}
res.push_back(t);
}
return res;
}
};
python3代码如下，
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def closestNodes(self, root: Optional[TreeNode], queries: List[int]) -> List[List[int]]:
nums = []
def dfs(node: TreeNode) -> None:
if node is None:
return
dfs(node.left)
nums.append(node.val)
dfs(node.right)
return
dfs(root)
res = []
for query in queries:
t = [-1, -1]
idx1 = bisect.bisect_left(nums, query+1)
idx1 -= 1
if 0 <= idx1 < len(nums):
t[0] = nums[idx1]
idx2 = bisect.bisect_left(nums, query)
if 0 <= idx2 < len(nums):
t[1] = nums[idx2]
res.append(t)
return res
题目71：74. 搜索二维矩阵
解题思路：二分，模拟bisect_left。
C++代码如下，
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int n = matrix.size();
int m = matrix[0].size();
int left = 0;
int right = n * m - 1;
int res = -1;
function<bool(int)> check =[&](int mid) ->bool {
int i = mid / m;
int j = mid % m;
return matrix[i][j] >= target;
};
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
if (res == -1) return false;
else {
int i = res / m;
int j = res % m;
return matrix[i][j] == target;
}
}
};
python3代码如下，
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
n = len(matrix)
m = len(matrix[0])
left = 0
right = n * m - 1
res = -1
def check(mid: int) -> bool:
i = mid // m
j = mid % m
return matrix[i][j] >= target
#模拟bisect_left
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1
else:
left = mid + 1
if res == -1:
return False
else:
i = res // m
j = res % m
return matrix[i][j] == target
题目72：240. 搜索二维矩阵 II
解题思路：二分。
C++代码如下，
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int n = matrix.size();
int m = matrix[0].size();
int i = n-1;
int j = 0;
while (i >= 0 && j < m) {
if (matrix[i][j] == target) {
return true;
} else if (matrix[i][j] < target) {
j += 1;
} else {
i -= 1;
}
}
return false;
}
};
python3代码如下，
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
n = len(matrix)
m = len(matrix[0])
i = n - 1
j = 0
while i >= 0 and j < m:
if matrix[i][j] == target:
return True
elif matrix[i][j] < target:
j += 1
else:
i -= 1
return False
题目73：278. 第一个错误的版本
解题思路：二分。
C++代码如下，
// The API isBadVersion is defined for you.
// bool isBadVersion(int version);
class Solution {
public:
int firstBadVersion(int n) {
int left = 1;
int right = n;
int res = -1;
while (left <= right) {
int mid = (1ll * left + right) / 2;
if (isBadVersion(mid)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
};
python3代码如下，
# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:
class Solution:
def firstBadVersion(self, n: int) -> int:
left = 1
right = n
res = -1
while left <= right:
mid = (left + right) // 2
if isBadVersion(mid):
res = mid
right = mid - 1
else:
left = mid + 1
return res
题目74：374. 猜数字大小
解题思路：二分。
C++代码如下，
/**
* Forward declaration of guess API.
* @param num your guess
* @return -1 if num is higher than the picked number
* 1 if num is lower than the picked number
* otherwise return 0
* int guess(int num);
*/
class Solution {
public:
int guessNumber(int n) {
int left = 1;
int right = n;
while (left <= right) {
int mid = (1ll * left + right) / 2;
int x = guess(mid);
if (x == 0) return mid;
else if (x == -1) right = mid - 1;
else left = mid + 1;
}
return -100;
}
};
python3代码如下，
# The guess API is already defined for you.
# @param num, your guess
# @return -1 if num is higher than the picked number
# 1 if num is lower than the picked number
# otherwise return 0
# def guess(num: int) -> int:
class Solution:
def guessNumber(self, n: int) -> int:
left = 1
right = n
while left <= right:
mid = (left + right) // 2
x = guess(mid)
if x == 0:
return mid
elif x == -1:
right = mid - 1
else:
left = mid + 1
return -100
题目75：162. 寻找峰值
解题思路：二分。
C++代码如下，
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int n = nums.size();
int left = 0;
int right = n - 1;
int res = -1;
function<bool(int)> check =[&] (int x) -> bool {
if (x-1 >= 0) return nums[x-1] < nums[x];
else return true;
};
while (left <= right) {
int mid = (1ll * left + right) / 2;
if (check(mid)) {
res = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def findPeakElement(self, nums: List[int]) -> int:
n = len(nums)
left = 0
right = n-1
res = -1
def check(mid: int) -> bool:
if mid-1 >= 0:
return nums[mid-1] < nums[mid]
else:
return True
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid #求最大值
left = mid + 1
else:
right = mid - 1
return res
题目76：1901. 寻找峰值 II
解题思路：转换思维，二分。
C++代码如下，
class Solution {
public:
vector<int> findPeakGrid(vector<vector<int>>& mat) {
int n = mat.size();
int m = mat[0].size();
function<bool(int)> check =[&] (int i) -> bool {
int mj = 0;
for (int j = 1; j < m; ++j) {
if (mat[i][j] > mat[i][mj]) {
mj = j;
}
}
return i-1 == -1 || mat[i-1][mj] <= mat[i][mj];
};
int resi = -1;
int left = 0;
int right = n-1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
resi = mid; //求最最大值
left = mid + 1;
} else {
right = mid - 1;
}
}
int resj = 0;
for (int j = 1; j < m; ++j) {
if (mat[resi][j] > mat[resi][resj]) {
resj = j;
}
}
return {resi, resj};
}
};
python3代码如下，
class Solution:
def findPeakGrid(self, mat: List[List[int]]) -> List[int]:
#二分每一行的最大值
n = len(mat)
m = len(mat[0])
left = 0
right = n - 1
resi = -1
def check(i: int) -> bool:
my = max(mat[i])
j = mat[i].index(my)
return i-1 == -1 or mat[i-1][j] <= mat[i][j]
while left <= right:
mid = (left + right) // 2
if check(mid):
resi = mid
left = mid + 1
else:
right = mid - 1
my = max(mat[resi])
resj = mat[resi].index(my)
return [resi, resj]
题目77：852. 山脉数组的峰顶索引
解题思路：二分。
C++代码如下，
class Solution {
public:
int peakIndexInMountainArray(vector<int>& arr) {
int n = arr.size();
int left = 0;
int right = n-1;
int res = -1;
function<bool(int)> check =[&] (int mid) -> bool {
return mid-1 == -1 || arr[mid-1] <= arr[mid];
};
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
left = mid + 1; //最大值
} else {
right = mid - 1;
}
}
return res;
}
};
python3代码如下，
class Solution:
def peakIndexInMountainArray(self, arr: List[int]) -> int:
n = len(arr)
left = 0
right = n-1
res = -1
def check(mid: int) -> bool:
return mid-1 == -1 or arr[mid-1] < arr[mid]
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
left = mid + 1 #求最大值
else:
right = mid - 1
return res
题目78：1095. 山脉数组中查找目标值
解题思路：二分。
C++代码如下，
/**
* // This is the MountainArray's API interface.
* // You should not implement it, or speculate about its implementation
* class MountainArray {
* public:
* int get(int index);
* int length();
* };
*/
class Solution {
public:
int findInMountainArray(int target, MountainArray &mountainArr) {
//求峰顶元素下标
int peek = -1;
int left = 0;
int right = mountainArr.length() - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (mid-1 == -1 || mountainArr.get(mid-1) < mountainArr.get(mid)) {
peek = mid;
left = mid + 1; //求最大值
} else {
right = mid - 1;
}
}
//[0,peek]求目标值target
left = 0;
right = peek;
int idx1 = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (mountainArr.get(mid) <= target) {
idx1 = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
if (idx1 != -1 && mountainArr.get(idx1) == target) {
return idx1;
}
//[peek,n-1]求目标值target
left = peek;
right = mountainArr.length() - 1;
int idx2 = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (mountainArr.get(mid) <= target) {
idx2 = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
if (idx2 != -1 && mountainArr.get(idx2) == target) {
return idx2;
}
return -1;
}
};
python3代码如下，
# """
# This is MountainArray's API interface.
# You should not implement it, or speculate about its implementation
# """
#class MountainArray:
# def get(self, index: int) -> int:
# def length(self) -> int:
class Solution:
def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
peek = -1 #顶峰坐标
left = 0
right = mountain_arr.length()-1
def check(mid: int) -> bool:
return mid-1 == -1 or mountain_arr.get(mid-1) <= mountain_arr.get(mid)
while left <= right:
mid = (left + right) // 2
if check(mid):
peek = mid #求最大值
left = mid + 1
else:
right = mid - 1
#[0,peek]求元素
idx1 = -1
left = 0
right = peek
while left <= right:
mid = (left + right) // 2
if mountain_arr.get(mid) <= target:
idx1 = mid
left = mid + 1
else:
right = mid - 1
if idx1 != -1 and mountain_arr.get(idx1) == target:
return idx1
#[peek,n-1]求元素
idx2 = -1
left = peek
right = mountain_arr.length()-1
while left <= right:
mid = (left + right) // 2
if mountain_arr.get(mid) <= target:
idx2 = mid
right = mid - 1
else:
left = mid + 1
if idx2 != -1 and mountain_arr.get(idx2) == target:
return idx2
return -1
题目79：153. 寻找旋转排序数组中的最小值
解题思路：小技巧，二分。
C++代码如下，
class Solution {
public:
int findMin(vector<int>& nums) {
int n = nums.size();
int left = 0;
int right = n - 1;
int idx = -1; //转折点
function<bool(int)> check =[&] (int mid) -> bool {
return (mid-1 == -1 || nums[mid-1] <= nums[mid]) && nums[0] <= nums[mid];
};
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
idx = mid;
left = mid + 1; //求最大值
} else {
right = mid - 1;
}
}
return nums[(idx+1)%n];
}
};
python3代码如下，
class Solution:
def findMin(self, nums: List[int]) -> int:
#找到peek元素
peek = -1
n = len(nums)
left = 0
right = n-1
while left <= right:
mid = (left + right) // 2
if (mid-1 == -1 or nums[mid-1] <= nums[mid]) and nums[0] <= nums[mid]:
peek = mid
left = mid + 1 #求最大值
else:
right = mid - 1
#print(f"peek = {peek}.")
return nums[(peek+1) % n]
题目80：33. 搜索旋转排序数组
解题思路：二分。
C++代码如下，
class Solution {
public:
int search(vector<int>& nums, int target) {
int n = nums.size();
int left = 0;
int right = n-1;
int idx = -1;
while (left <= right) {
int mid = (1ll * left + right) / 2;
if ((mid-1 == -1 || nums[mid-1] <= nums[mid]) && nums[0] <= nums[mid]) {
idx = mid;
left = mid + 1; //求最大值
} else {
right = mid - 1;
}
}
//[0,idx]求目标值target
left = 0;
right = idx;
int ans1 = -1;
while (left <= right) {
int mid = (1ll * left + right) / 2;
if (nums[mid] <= target) {
ans1 = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
if (ans1 != -1 && nums[ans1] == target) {
return ans1;
}
//[idx+1,n-1]求目标值target
left = idx + 1;
right = n - 1;
int ans2 = -1;
while (left <= right) {
int mid = (1ll * left + right) / 2;
if (nums[mid] <= target) {
ans2 = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
if (ans2 != -1 && nums[ans2] == target) {
return ans2;
}
return -1;
}
};
python3代码如下，
class Solution:
def search(self, nums: List[int], target: int) -> int:
#找到转折点
n = len(nums)
left = 0
right = n - 1
idx = -1
def check(mid: int) -> bool:
return (mid-1 == -1 or nums[mid-1] < nums[mid]) and nums[0] <= nums[mid]
while left <= right:
mid = (left + right) // 2
if check(mid):
idx = mid
left = mid + 1
else:
right = mid - 1
#print(f"idx = {idx}.")
#[0,idx]找目标值target
ans1 = -1
left = 0
right = idx
#print(f"step1: left = {left}, right = {right}.")
while left <= right:
mid = (left + right) // 2
if nums[mid] <= target:
ans1 = mid
left = mid + 1
else:
right = mid - 1
if ans1 != -1 and nums[ans1] == target:
return ans1
#[idx+1,n-1]找目标值target
ans2 = -1
left = idx + 1
right = n - 1
#print(f"step2: left = {left}, right = {right}.")
while left <= right:
mid = (left + right) // 2
if nums[mid] <= target:
ans2 = mid
left = mid + 1
else:
right = mid - 1
if ans2 != -1 and nums[ans2] == target:
return ans2
return -1
题目81：222. 完全二叉树的节点个数
解题思路：简单题简单做。
C++代码如下，
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
int cnt = 0;
function<void(TreeNode* node)> dfs =[&] (TreeNode* node) -> void {
if (node == nullptr) return;
dfs(node->left);
cnt += 1;
dfs(node->right);
return;
};
dfs(root);
return cnt;
}
};
python3代码如下，
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def countNodes(self, root: Optional[TreeNode]) -> int:
#简单题简单做
cnt = 0
def dfs(node: TreeNode) -> None:
nonlocal cnt
if node is None:
return
dfs(node.left)
cnt += 1
dfs(node.right)
return
dfs(root)
return cnt
题目82：1539. 第 k 个缺失的正整数
解题思路：简单题简单做。
C++代码如下，
class Solution {
public:
int findKthPositive(vector<int>& arr, int k) {
set<int> brr(arr.begin(), arr.end());
int cnt = 0;
int i = 1;
while (true) {
if (brr.find(i) == brr.end()) {
cnt += 1;
if (cnt == k) {
return i;
}
}
i += 1;
}
return -100;
}
};
python3代码如下，
class Solution:
def findKthPositive(self, arr: List[int], k: int) -> int:
#简单题简单做
arr = set(arr)
i = 1
cnt = 0
while True:
if i not in arr:
cnt += 1
if cnt == k:
return i
i += 1
return -100
题目83：540. 有序数组中的单一元素
解题思路：二分。
C++代码如下，
class Solution {
public:
int singleNonDuplicate(vector<int>& a) {
int n = a.size();
if (n == 1) { //特判
return a[0];
}
int left = 0;
int right = n - 1;
int res = -1;
function<bool(int)> check =[&] (int mid) -> bool {
if (mid % 2 == 0) {
return (mid+1 < n && a[mid] != a[mid+1]) || (mid+1 == n && a[mid] == a[mid-1]);
} else {
return a[mid] == a[mid+1];
}
};
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1; //求最小值
} else {
left = mid + 1;
}
}
if (res == -1) {
res = n - 1; //特判[1,1,2]这种情况
}
return a[res];
}
};
python3代码如下，
class Solution:
def singleNonDuplicate(self, a: List[int]) -> int:
n = len(a)
left = 0
right = n - 1
res = -1
def check(mid: int) -> bool:
if mid % 2 == 0:
return (mid+1 < n and a[mid] != a[mid+1]) or (mid+1 == n and a[mid] == a[mid-1])
else:
return a[mid] == a[mid+1]
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1 #求最小值
else:
left = mid + 1
return a[res]
题目84：4. 寻找两个正序数组的中位数
解题思路：求第k小的数，二分。
C++代码如下，
class Solution {
public:
int compute_kth_num(const vector<int>& nums1, const vector<int>& nums2, const int k) {
//计算第k小的数
//小于等于mid的数的个数cnt,cnt>=k
//求mid的最小值
int n = nums1.size();
int m = nums2.size();
int left = 1e7;
int right = -1e7;
if (n > 0) {
left = min(left, nums1[0]);
right = max(right, nums1[n-1]);
}
if (m > 0) {
left = min(left, nums2[0]);
right = max(right, nums2[m-1]);
}
function<bool(int)> check =[&] (int mid) -> bool {
auto iter = lower_bound(nums1.begin(), nums1.end(), mid+1);
int idx1 = distance(nums1.begin(), iter);
iter = lower_bound(nums2.begin(), nums2.end(), mid+1);
int idx2 = distance(nums2.begin(), iter);
return idx1+idx2 >= k;
};
int res = -1;
while (left <= right) {
int mid = (left + right) / 2;
if (check(mid)) {
res = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
int m = nums2.size();
int k = (n + m) / 2 + 1;
if ((n+m) % 2 == 0) {
int k1 = k - 1;
int k2 = k;
int ans1 = compute_kth_num(nums1, nums2, k1);
int ans2 = compute_kth_num(nums1, nums2, k2);
double ans = (1.0 * ans1 + ans2) / 2.0;
return ans;
} else {
int ans = compute_kth_num(nums1, nums2, k);
return ans;
}
}
};
python3代码如下，
class Solution:
def compute_kth_num(self, nums1: List[int], nums2: List[int], k: int) -> int:
#print(f"nums1 = {nums1}, nums2 = {nums2}, k = {k}.")
#求第k小的数
n = len(nums1)
m = len(nums2)
left = int(1e7)
right = -1 * int(1e7)
if n > 0:
left = min(left, nums1[0])
right = max(right, nums1[-1])
if m > 0:
left = min(left, nums2[0])
right = max(right, nums2[-1])
def check(mid: int) -> bool:
idx1 = bisect_left(nums1, mid+1)
idx2 = bisect_left(nums2, mid+1)
return idx1+idx2 >= k
#计算<=mid的个数，它>=k
#求mid的最小值
res = -1
while left <= right:
mid = (left + right) // 2
if check(mid):
res = mid
right = mid - 1
else:
left = mid + 1
#print(f"res = {res}.")
return res
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
n = len(nums1)
m = len(nums2)
k = (n + m) // 2
#求第k小的数
if (n + m) % 2 == 0:
k1 = k
k2 = k + 1
ans1 = self.compute_kth_num(nums1,nums2,k1)
ans2 = self.compute_kth_num(nums1,nums2,k2)
ans = (ans1 + ans2) / 2
return ans
else:
ans = self.compute_kth_num(nums1, nums2, k + 1)
return ans
