# 单词搜索
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.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15board 和 word 仅由大小写英文字母组成
进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
## template
```java
class Solution {
public boolean exist(char[][] board, String word) {
int cl = board.length;
int rl = board[0].length;
boolean[][] flag = new boolean[cl][rl];
for (int i = 0; i < cl; i++) {
for (int j = 0; j < rl; j++) {
if (find(board, word, flag, i, j, 0))
return true;
}
}
return false;
}
public boolean find(char[][] board, String word, boolean[][] flag, int i, int j, int index) {
int cl = board.length;
int rl = board[0].length;
if (word.length() == index)
return true;
if (i < 0 || i >= cl || j >= rl || j < 0)
return false;
if (flag[i][j] || word.charAt(index) != board[i][j])
return false;
flag[i][j] = true;
boolean judge = find(board, word, flag, i - 1, j, index + 1) || find(board, word, flag, i + 1, j, index + 1)
|| find(board, word, flag, i, j - 1, index + 1) || find(board, word, flag, i, j + 1, index + 1);
flag[i][j] = false;
return judge;
}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```