【leetcode周赛记录—

405周赛记录

#1.leetcode100339_找出加密后的字符串
2.leetcode100328_生成不含相邻零的二进制字符串
3.leetcode100359_统计X和Y频数相等的子矩阵数量
4.leetcode100350_最小代价构造字符串

刷了一段时间算法了，打打周赛看看什么水平了

#1.leetcode100339_找出加密后的字符串

思路：感觉还是挺简单的，第一次还以为是字典序的后k位，审题不认真。。。
偷了个懒，忘记怎么原地修改数组了，感觉要用辅助空间，还是直接库函数吧
用后k位字符替换，其实就是把后k位提前嘛
1，2，… ，k-1，k，k+1，…，size
1+k，k，… ，size，1，2，…，k

	string getEncryptedString(string s, int k) {k = k % s.size();return s.substr(k, s.size()) + s.substr(0, k);}

2.leetcode100328_生成不含相邻零的二进制字符串

思路：回溯，如果当前是0，下一个就不是0，除此之外，下一个都可以是1

class Solution {vector<string> res;string path;
public:void backtracking(int n) {if(path.size() == n) {res.push_back(path);return;}if(path == "" || path.back() != '0') {path += '0';backtracking(n);path.pop_back();}path += '1';backtracking(n);path.pop_back();return;}vector<string> validStrings(int n) {res.clear();path.clear();backtracking(n);return res;}
};

3.leetcode100359_统计X和Y频数相等的子矩阵数量

思路：动态规划吧，dp[i] = dp[i-1][j-1] + (dp[i-1][j] - dp[i-1][j-1]) + (dp[i][j-1] - dp[i-1][j-1])
大概思路就是以示例1为例
dp[i][j]初始化为1; 更新为2

	// dp[i][j] 表示 从左上角开始长为i宽为j的矩形所含有的 x和 y的数量class Solution {
public:int numberOfSubmatrices(vector<vector<char>>& grid) {int m = grid.size();int n = grid[0].size();vector<vector<pair<int, int>>> dp(m, vector<pair<int, int>>(n, {0, 0}));if(grid[0][0] == 'X'){dp[0][0].first = 1;} else if(grid[0][0] == 'Y') {dp[0][0].second = 1;}for(int i = 1; i < m; i++){dp[i][0] = dp[i-1][0];if(grid[i][0] == 'X'){dp[i][0].first++;} else if(grid[i][0] == 'Y') {dp[i][0].second++;}}for(int j = 1; j < n; j++){dp[0][j] = dp[0][j-1];if(grid[0][j] == 'X'){dp[0][j].first++;} else if(grid[0][j] == 'Y') {dp[0][j].second++;}}//         cout << "1. === \n";//         for(int i = 0; i < m; i++) {
//             for(int j = 0; j < n; j++){
//                 cout << "(" << dp[i][j].first << "," << dp[i][j].second << "), ";
//             }
//             cout << endl;
//         }for(int i = 1; i < m; i++)for(int j = 1; j < n; j++) {dp[i][j].first = dp[i][j-1].first + dp[i-1][j].first - dp[i-1][j-1].first;dp[i][j].second = dp[i][j-1].second + dp[i-1][j].second - dp[i-1][j-1].second;if(grid[i][j] == 'X'){dp[i][j].first++;}else if(grid[i][j] == 'Y'){                    dp[i][j].second++;}}int res = 0;//         cout << "2. === \n";//         for(int i = 0; i < m; i++) {
//             for(int j = 0; j < n; j++){
//                 cout << "(" << dp[i][j].first << "," << dp[i][j].second << "), ";
//             }
//             cout << endl;
//         }for(int i = 0; i < m; i++)for(int j = 0; j < n; j++){if(dp[i][j].first != 0 && dp[i][j].first == dp[i][j].second)res++;  }return res;}
};