题解：模拟
代码:
class Solution {
public:
bool validUtf8(vector<int>& data) {
int bit = 0;
for(auto num : data){
if(bit == 0){
if((num >>5) == 0b110) bit = 1;
else if((num >>4) == 0b1110) bit = 2;
else if((num >>3) == 0b11110) bit = 3;
else if(num >>7) return false;
}else{
if((num >> 6) != 0b10) return false;
bit--;
}
}
return bit == 0;
}
};
题解：哈希表
代码:
class Solution {
public:
vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
map<string, int> s1;
for(int i=0; i<list1.size(); i++){
s1[list1[i]] = i;
}
map<string,int> s2;
for(int i=0; i<list2.size(); i++){
s2[list2[i]] = i;
}
set<string> con(list1.begin(), list1.end());
for(auto& s : list2){
con.insert(s);
}
vector<string> res;
int minv = INT_MAX;
for(auto&s : con){
if(s1.count(s) && s2.count(s)){
int diff = s1[s] + s2[s];
if( diff < minv){
minv = diff;
res.clear();
res.push_back(s);
}else if(minv == diff){
res.push_back(s);
}
}
}
return res;
}
};
题解：二进制枚举
代码:
class Solution {
public:
int countMaxOrSubsets(vector<int>& nums) {
int n = nums.size(), m = 1<<n;
int maxv = 0, res = 0;
for(int i=0; i<m; i++){
int u = 0;
for(int j=0; j<n; j++){
if((i>>j)&1==1){
u |= nums[j];
}
}
if(u > maxv){
maxv = u;
res = 1;
}else if(u == maxv){
res ++;
}
}
return res;
}
};
题解: 哈希表
代码:
class AllOne {
public:
unordered_map<string, int> f;
set<pair<int, string>> set;
AllOne() {
}
void inc(string key) {
int count = f[key]++;
if(count){
set.erase({count, key});
}
set.insert(make_pair(f[key], key));
}
void dec(string key) {
if(f[key] <= 0){
return;
}
int count = f[key]--;
set.erase({count, key});
if(f[key]){
set.insert({f[key], key});
}
}
string getMaxKey() {
if(set.empty()) return "";
return set.rbegin()->second;
}
string getMinKey() {
if(set.empty()) return "";
return set.begin()->second;
}
};
/**
* Your AllOne object will be instantiated and called as such:
* AllOne* obj = new AllOne();
* obj->inc(key);
* obj->dec(key);
* string param_3 = obj->getMaxKey();
* string param_4 = obj->getMinKey();
*/
题解: 字典树
代码:
const int N = 3e4+10;
const int M = 26;
class Solution {
public:
int f[N][M];
bool st[N];
int idx;
void add(string s){
int n = s.size();
int p = 0;
for(int i=0; i<n; i++){
if(f[p][s[i]-'a'] == 0) f[p][s[i]-'a'] = ++idx;
p = f[p][s[i]-'a'];
}
st[p] = true;
}
bool query(string s){
int n = s.size();
int p = 0;
for(int i=0; i<n; i++){
p = f[p][s[i]-'a'];
if(!st[p]) return false;
}
return true;
}
string longestWord(vector<string>& words) {
memset(f, 0, sizeof f);
memset(st, false, sizeof st);
idx = 0;
for(auto& s : words){
add(s);
}
string res;
for(auto& s : words){
if(s.size() < res.size()) continue;
if(s.size() == res.size() && s > res){
continue;
}
if(query(s)) res = s;
}
return res;
}
};
题解: 哈希表
代码:
class Bank {
public:
unordered_map<int, long long> map;
Bank(vector<long long>& balance) {
for(int i=1; i<=balance.size(); i++){
map[i-1] = balance[i-1];
}
}
bool transfer(int account1, int account2, long long money) {
if(account1 <= 0 || account1 > map.size()) return false;
if(account2 <= 0 || account2 > map.size()) return false;
if(map[account1-1] >= money){
map[account1-1] -= money;
map[account2-1] += money;
return true;
}
return false;
}
bool deposit(int account, long long money) {
if(account <= 0 || account > map.size()) return false;
map[account-1] += money;
return true;
}
bool withdraw(int account, long long money) {
if(account <= 0 || account > map.size()) return false;
if(map[account-1] < money) return false;
map[account-1] -= money;
return true;
}
};
/**
* Your Bank object will be instantiated and called as such:
* Bank* obj = new Bank(balance);
* bool param_1 = obj->transfer(account1,account2,money);
* bool param_2 = obj->deposit(account,money);
* bool param_3 = obj->withdraw(account,money);
*/
题解: 深度搜索
代码:
/**
* 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:
string dfs(TreeNode* root){
if(!root) return "";
else{
string res = to_string(root->val);
if(root->left){
string left = dfs(root->left);
res += "("+left+")";
}
if(!root->left && root->right){
res += "()";
}
if(root->right){
string right = dfs(root->right);
res += "("+right+")";
}
return res;
}
}
string tree2str(TreeNode* root) {
return dfs(root);
}
};
