来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/word-search

题目描述：

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

思路：

因为没有给定起点，所以最开始需要遍历二维数组，寻找起始点。board[start][end] == word的第一个单词的话，就以start,end为起点开始回溯遍历。递归结束还没有找到说明不是这个起点，继续寻找board[start][end]=word.charAt(0)的节点，重复刚才的操作。

79,单词搜索.png

实现：

class Solution {
    int m = 0;
    int n = 0;

    boolean flag = false;
    public boolean exist(char[][] board, String word) {
        m = board.length;
        n = board[0].length;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (flag) {
                    return flag;
                }
                if (word.charAt(0) == board[i][j]) {
                    dfs(board, word, 0, i, j);
                }
            }
        }
        return flag;
    }
    public void dfs(char[][] board, String word, int index, int start, int end) {
         if (index == word.length()) {
            flag = true;
            return;
        }
        if (start < 0 || end < 0 || start >= m || end >= n) {
            return;
        }
        if (board[start][end] == '*') {
            return;
        }
      
        if (word.charAt(index) == board[start][end]) {
            char temp = board[start][end];
            board[start][end] = '*';
            dfs(board, word, index + 1, start + 1, end);
            dfs(board, word, index + 1, start - 1, end);
            dfs(board, word, index + 1, start, end + 1);
            dfs(board, word, index + 1, start, end - 1);
            board[start][end] = temp;
        }
       
    }
}