LeetCode--079--单词搜索(python)

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例:

board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

给定 word = "ABCCED", 返回 true.
给定 word = "SEE", 返回 true.
给定 word = "ABCB", 返回 false.

 class Solution:
     #         (x-1,y)
     # (x,y-1) (x,y) (x,y+1)
     #         (x+1,y)
     directions= [(0,-1),(-1,0),(0,1),(1,0)]
     def exist(self, board: List[List[str]], word: str) -> bool:
         m = len(board)
         if m == 0:
             return False
         n = len(board[0])

         marked = [[False for _ in range(n)] for _ in range(m)]
         # 对每一个格子都从头开始搜索
         for i in range(m):
             for j in range(n):
                 if self.__search_word(board,word,0,i,j,marked,m,n):
                     return True
         return False
     def  __search_word(self,board,word,index,start_x,start_y,marked,m,n):
         # 先写递归终止条件
         if index == len(word) -1:
             return board[start_x][start_y] == word[index]
         # 中间匹配了，再继续搜索
         if board[start_x][start_y] == word[index]:
             # 先占住这个位置，搜索不成功的话，要释放掉
             marked[start_x][start_y] = True
             for direction in self.directions:
                 new_x = start_x +direction[0]
                 new_y = start_y + direction[1]
                 ## 注意：如果这一次 search word 成功的话，就返回
                 if 0 <= new_x < m and 0 <= new_y < n and not marked[new_x][new_y] and self.__search_word(board,word,index+1,new_x,new_y,marked,m,n):
                     return True
             marked[start_x][start_y] = False
         return False