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/video/av25763384?t=813

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 True
        if x < 0 or x == len(board) or y < 0 or y == len(board[0]):
            return False
        if board[x][y] != word[index]:
            return False
        source = 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] = source
        return exist

    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        if len(word) == 0:
            return False
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.dfs(board, word, 0, i, j):
                    return True
        return False

[Leetcode] 79.单词搜索

题目

题解

java

python

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