题目

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.

题解

这个题目拿到题目就应该能想到是用DFS的题目，因为这完完全全就是DFS，没有做任何的变形，关于DFS，这里就不重复讲解。

推荐一个b站上的视频，不熟悉的同学可以回顾一下。

https://www.bilibili.com/vide...

熟悉的同学直接看代码吧

java

class Solution {public boolean exist(char[][] board, String word) {if (word == null || word.length() == 0) {return true;}char[] chs = word.toCharArray();for (int i = 0; i < board.length; i++) {for (int j = 0; j < board[0].length; j++) {if (dfs(board, chs, 0, i, j)) {return true;}}}return false;}private boolean dfs(char[][] board, char[] words, int index, int x, int y) {if (index == words.length) {return true;}if (x < 0 || x == board.length || y < 0 || y == board[0].length) {return false;}if (board[x][y] != words[index]) {return false;}char source = board[x][y];board[x][y] = '\0';boolean exist = dfs(board, words, index + 1, x, y + 1)|| dfs(board, words, index + 1, x, y - 1)|| dfs(board, words, index + 1, x + 1, y)|| dfs(board, words, index + 1, x - 1, y);board[x][y] = source;return exist;}
}

python

class Solution:def dfs(self, board, word, index, x, y):if not board or index == len(word):return Trueif x < 0 or x == len(board) or y < 0 or y == len(board[0]):return Falseif board[x][y] != word[index]:return Falsesource = board[x][y]board[x][y] = '\0'exist = self.dfs(board, word, index + 1, x, y + 1) or self.dfs(board, word, index + 1, x, y - 1) or self.dfs(board, word, index + 1, x + 1, y) or self.dfs(board, word, index + 1, x - 1, y)board[x][y] = sourcereturn existdef exist(self, board, word):""":type board: List[List[str]]:type word: str:rtype: bool"""if len(word) == 0:return Falsefor i in range(len(board)):for j in range(len(board[0])):if self.dfs(board, word, 0, i, j):return Truereturn False

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