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