120 Triangle Medium
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
我的方法：
从第二行开始计算，每一行下标loc的数 = 自身 + min(上一行下标loc-1的数，上一行下标loc的数)，loc-1与loc需要判断在上一行是否越界。
最后一行的最小值就是结果。
class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if(triangle == null || triangle.size() == 0 || triangle.get(0).size() == 0) {
return 0;
}
List<List<Integer>> status = new ArrayList<>();
List<Integer> first = new ArrayList<>();
first.add(triangle.get(0).get(0));
status.add(first);
for(int i=1; i<triangle.size(); i++) {
List<Integer> line = new ArrayList<>();
int j=0;
while(j<i+1) {
int left = Integer.MAX_VALUE;
int right = Integer.MAX_VALUE;
if(j-1>=0) {
left = status.get(i-1).get(j-1);
}
if(j<status.get(i-1).size()) {
right = status.get(i-1).get(j);
}
line.add(triangle.get(i).get(j) + Math.min(left, right));
j++;
}
status.add(line);
}
Object[] lastLine = status.get(triangle.size()-1).toArray();
Arrays.sort(lastLine);
return (int)lastLine[0];
}
}
更好的方法：
反过来，从倒数第二行开始计算。
f(i,j) 为从位置 (i,j) 出发，路径的最小和，则 f(i,j) = min{f(i+1,j),f(i+1,j+1)} + (i,j)
最后求到f(0,0)就是结果。
class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if(triangle == null || triangle.size() == 0 || triangle.get(0).size() == 0) {
return 0;
}
for(int i= triangle.size()-2; i>=0; i--) {
for(int j=0; j<i+1; j++) {
int cur = triangle.get(i).get(j);
triangle.get(i).set(j, cur + Math.min(triangle.get(i+1).get(j), triangle.get(i+1).get(j+1)));
}
}
return triangle.get(0).get(0);
}
}
53 Maximum Subarray Easy
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
经典题：最大连续子序列和
从头到尾依次遍历数组元素，对每一个元素，有两种选择：加入之前的SubArray，自己另起一个SubArray。
设状态为f[j]，表示以S[j]结尾的最大连续子序列和，则状态转移方程：
f[j] = max{f[j-1]+S[j], S[j]}, i≤j≤n
target = max{f[j]}, i≤j≤n
class Solution {
public int maxSubArray(int[] nums) {
if(nums == null || nums.length == 0) {
return 0;
}
int result = Integer.MIN_VALUE;
int preSum = 0;
for(int i=0; i<nums.length; i++) {
preSum = Math.max(preSum + nums[i], nums[i]);
result = Math.max(preSum, result);
}
return result;
}
}
375 Guess Number Higher or Lower II Medium
We are playing the Guess Game. The game is as follows: I pick a number from 1 to n. You have to guess which number I picked. Every time you guess wrong, I'll tell you whether the number I picked is higher or lower. However, when you guess a particular number x, and you guess wrong, you pay x dollars. You win the game when you guess the number I picked. Example: n = 10, I pick 8. First round: You guess 5, I tell you that it's higher. You pay 5 dollars. Second round: You guess 7, I tell you that it's higher. You pay 7 dollars. Third round: You guess 9, I tell you that it's lower. You pay 9 dollars. Game over. 8 is the number I picked. You end up paying 5 + 7 + 9 = 21. Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.
参考：www.hrwhisper.me/leetcode-gu…
62 Unique Paths Medium
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
- Right -> Right -> Down
- Right -> Down -> Right
- Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
f(i,j)表示从起点(i,j)到达(i,j)的路线数
状态转移方程：f(i,j) = f(i-1,j)+f(i,j-1)
class Solution {
public int uniquePaths(int m, int n) {
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i < m + 1; i++) {
for (int j = 1; j < n + 1; j++) {
if (i == 1 && j == 1) {
dp[i][j] = 1;
continue;
}
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m][n];
}
}
