【LeetCode】79.单词搜索

题目

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

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 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

解答

源代码

class Solution {
    public boolean exist(char[][] board, String word) {
        // 表示当前格子是否被经过
        boolean visited[][] = new boolean[board.length][board[0].length];

        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j = 0 && newj >= 0 && newi 

总结

这个题目还挺熟悉的，应该是大一小学期实训的时候做过，那时候做的是迷宫问题，都是方向+回溯。