79：单词搜索

问题描述

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.

思路

思路倒是很简单。用DFS找出来符合条件的即可。不过我可能是老了，竟然不会写DFS了。真的蛋疼。

AC代码

class Solution:
    def exist(self, board, word: str) -> bool:
        target_word = word[0]
        self.height = len(board)
        self.width = len(board[0])
        self.used = [[False for i in range(self.width)] for i in range(self.height)]
        flag = False
        for i in range(self.height):
            for j in range(self.width):
                if board[i][j] == target_word:
                    self.used[i][j] = True
                    flag = self.dfs(board,i,j,word[1:])
                    if flag:
                        return flag
                    self.used[i][j] = False # 状态回退
        return flag
    def dfs(self,board,i,j,word):
        if len(word) == 0:
            return True
        target_search = word[0]
        stack = [] # 搜索四个方向。stack暂存结果
        if 0<j and not self.used[i][j-1] and board[i][j-1] == target_search:
            stack.append((i,j-1))
        if j<self.width-1 and not self.used[i][j+1] and board[i][j+1] == target_search:
            stack.append((i,j+1))
        if 0<i and not self.used[i-1][j] and board[i-1][j] == target_search:
            stack.append((i-1,j))
        if i<self.height-1 and not self.used[i+1][j] and board[i+1][j] == target_search:
            stack.append((i+1,j))
        for i in stack:
            self.used[i[0]][i[1]] = True
            flag = self.dfs(board,i[0],i[1],word[1:])
            if flag:
                return flag
            self.used[i[0]][i[1]] = False # 状态回退
        return False

s = Solution()
print(s.exist([["C","A","A"],["A","A","A"],["B","C","D"]]
,"AAB"))