题解 | #单词搜索#

利用 dfs 进行搜索，避免重复访问加入辅助数组 vistited，注意在每次路径递归失败后,将辅助数组的值置回初始值。

#
# 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 board string字符串一维数组 
# @param word string字符串 
# @return bool布尔型
#
class Solution:
    def exist(self , board: List[str], word: str) -> bool:
        # write code here
        visited = [[False for _ in range(len(board[0]))] for _ in range(len(board))]
        for i in range(len(board)):
            for j in range(len(board[i])):
                if board[i][j] == word[0]:
                    if self.search(i, j, len(board), len(board[i]), visited, board, word[1:]):
                        return True
        return False
                    
    def search(self, x, y, m, n, visited, board, word):
        if not word:
            return True
        visited[x][y] = True
        direation = [(-1, 0), (0, 1), (1, 0), (0, -1)]
        for i, j in direation:
            xi, yi = x + i, y + j
            if 0 <= xi < m and 0 <= yi < n and not visited[xi][yi] and board[xi][yi] == word[0]:
                if self.search(xi, yi, m, n, visited, board, word[1:]):
                    return True
        visited[x][y] = False
        return False