Prove safety offer 12: path matrix

The meaning of problems

Please design a function that determines whether there is a route that contains all the characters of a character string in a matrix. A lattice path may start from any of the matrix, each step in the matrix can be left, right, up, down one grid. If a route that passes through a matrix of a grid, the path can not enter the lattice. E.g

For example

the matrix containing the path "bcced" in a string, but does not contain the matrix "abcb" path, as a character string in the first row of the matrix b occupy the first second after the lattice, can not re-enter the path this grid.

Thinking

class Solution {
public:
    bool core(char *matrix, int row, int col, int rows, int cols, char *str, bool **visit) {
        //这是个递归调用，传入到最后了，说明整个字符串检测完了，返回true
        if (*str == 0)
            return true;
        //如果当前要访问的格子超过边界，或者不相等，或者已经被访问过了，都要返回false
        if (row >= rows || col >= cols || row < 0 || col < 0 || matrix[col + row * cols] != *str || visit[row][col])
            return false;
        bool has_path = false;
        //假设当前格子被访问
        visit[row][col] = 1;
        
        //递归的思想，从当前格子出发，上下左右是否有路径
        has_path = core(matrix, row + 1, col, rows, cols, str + 1, visit) ||
                   core(matrix, row, col + 1, rows, cols, str + 1, visit) ||
                   core(matrix, row - 1, col, rows, cols, str + 1, visit) ||
                   core(matrix, row, col - 1, rows, cols, str + 1, visit);
        
        //如果没有路径,就回退,然后这个格子变成没访问过的
        if (!has_path)
            visit[row][col] = 0;
        //返回
        return has_path;
    }

    bool hasPath(char *matrix, int rows, int cols, char *str) {
        //设计一个visit数组，用来标记已经访问过的单元
        bool **visit = new bool *[rows];
        for (int i = 0; i < rows; i++)
            visit[i] = new bool[cols];
        for (int i = 0; i < rows; i++)
            for (int j = 0; j < cols; j++)
                visit[i][j] = 0;
        //从任意单元格开始检测是否存在指定字符串
        for (int i = 0; i < rows; i++)
            for (int j = 0; j < cols; j++) {
                if (core(matrix, i, j, rows, cols, str, visit))
                    return true;
            }
        return false;
    }
};