题解 | #矩阵中的路径#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param matrix char字符型二维数组 
# @param word string字符串 
# @return bool布尔型
#
class Solution:
    def dfs(self, m,n, matrix, visited, word, k):

        if k >= len(word):
            return True
        if m < 0 or n < 0 or m >= len(matrix) or n >= len(matrix[0]) or visited[m][n] == 1:
            return False
        print(m, n, k, word[k])

        if matrix[m][n] == word[k]:
            visited[m][n] = 1
            if self.dfs(m+1,n,matrix,visited, word, k+1) or self.dfs(m-1,n,matrix,visited, word, k+1) or self.dfs(m,n+1,matrix,visited, word, k+1) or self.dfs(m,n-1,matrix,visited, word, k+1):
                    return True
            else:
                visited[m][n] = 0
                return False
        return False

    def hasPath(self , matrix , word ):
        # write code here
        l = len(matrix)
        w = len(matrix[0])
        for i in range(l):
            for j in range(w):
                visited = multilist = [[0 for col in range(w)] for row in range(l)]
                
                if matrix[i][j] == word[0]:                  
                    res = self.dfs(i,j, matrix, visited, word, 0)
#                     print(matrix[i][j], res)
                    if res:
                        return True
        return False