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.

思路：搜索回溯基本上是经典模板题了。

class Solution {private boolean[][] marked;//        x-1,y// x,y-1  x,y    x,y+1//        x+1,yprivate int[][] direction = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};// 盘面上有多少行private int m;// 盘面上有多少列private int n;private String word;private char[][] board;public boolean exist(char[][] board, String word) {m = board.length;if (m == 0)return false;n = board[0].length;marked = new boolean[m][n];this.word = word;this.board = board;for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (dfs(i, j, 0))return true;return false;}private boolean dfs(int i, int j, int start) {if (start == word.length() - 1) {return board[i][j] == word.charAt(start);}if (board[i][j] == word.charAt(start)) {marked[i][j] = true;for (int k = 0; k < 4; k++) {int newX = i + direction[k][0];int newY = j + direction[k][1];if (newX >= 0 && newX < m && newY >= 0 && newY < n && !marked[newX][newY]) {if (dfs(newX, newY, start + 1)) {return true;}}}marked[i][j] = false;}return false;}
}