https://www.nowcoder.com/practice/c61c6999eecb4b8f88a98f66b273a3cc?tpId=13&tqId=11218&tPage=4&rp=4&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一个格子开始，每一步可以在矩阵中向左，向右，向上，向下移动一个格子。如果一条路径经过了矩阵中的某一个格子，则该路径不能再进入该格子。

https://leetcode.cn/problems/ji-qi-ren-de-yun-dong-fan-wei-lcof/description/

class Solution {public:bool hasPath(char* matrix, int rows, int cols, char* str, int path_len, int i,int j, vector<vector<bool>>& visited) {if (str[path_len] == '\0') {return true;}if (i < 0 || i >= rows || j < 0 || j >= cols || visited[i][j] ||matrix[i * cols + j] != str[path_len]) {return false;}visited[i][j] = true;path_len++;bool res =  hasPath(matrix, rows, cols, str, path_len, i + 1, j, visited) ||hasPath(matrix, rows, cols, str, path_len, i - 1, j, visited) ||hasPath(matrix, rows, cols, str, path_len, i, j + 1, visited) ||hasPath(matrix, rows, cols, str, path_len, i, j - 1, visited);if (!res) {visited[i][j] = false;path_len--;}/* 注意，不加上面条件也可以，可以这么理解，如果是true，就直接返回了。visited[i][j] = false;path_len--;*/return res;}bool hasPath(char* matrix, int rows, int cols, char* str) {// 11:10vector<vector<bool>> visited(rows, vector<bool>(cols, false));int path_len = 0;for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; j++) {bool res = hasPath(matrix, rows, cols, str, path_len, i, j, visited);if (res) {return true;}}}return false;}};