矩阵中的路径python

思路：和机器人的运动范围类似
运用回溯法和DFS方法进行遍历比较

• 遍历整个矩阵，找到和字符串匹配的第一个字符

• 然后遍历该字符的上下左右字符，如有匹配，就把该匹配字符作为节点，继续匹配其上下左右字符，如不匹配，则退回到上个字符，继续匹配

• 为了防止重复，需要一个矩阵来存储走过的字符位置

注意： 本题给定的是一个字符串，在对字符串（一维）进行索引遍历的时候，应i,j表示矩阵索引，则字符串索引应该为 i * cols + j

代码：

# -*- coding:utf-8 -*-
class Solution:
    def find_path(self, matrix, rows, cols, path, path_number, i, j, boolean_if_go):
        has_next_path = False
        if len(path) == path_number:
            return True
        if i < rows and j < cols and i >= 0 and j >= 0 and matrix[i * cols + j] == path[path_number] and\
                not boolean_if_go[i * cols + j]:
            path_number += 1
            boolean_if_go[i * cols + j] = True
            has_next_path = (self.find_path(matrix, rows, cols, path, path_number, i-1, j, boolean_if_go)
                             or self.find_path(matrix, rows, cols, path, path_number, i, j+1, boolean_if_go)
                             or self.find_path(matrix, rows, cols, path, path_number, i+1, j, boolean_if_go)
                             or self.find_path(matrix, rows, cols, path, path_number, i, j-1, boolean_if_go))
            if not has_next_path:
                boolean_if_go[i * cols +j] = False
                path_number -= 1
        return has_next_path

    def hasPath(self, matrix, rows, cols, path):
        # write code here
        boolean_if_go = [0] * (rows * cols)
        path_number = 0
        for i in range(rows):
            for j in range(cols):
                if self.find_path(matrix, rows, cols, path, path_number, i, j, boolean_if_go):
                    return True
        return False