问题：

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[['A','B','C','E'],['S','F','C','S'],['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

解决：

【题意】题目要求在字母矩阵中，可以上下左右移动，看是否可以得到给定的字符串，同一位置的字符串只能使用一次。

① 使用dfs遍历。向上下左右四个方向移动。

class Solution{//10ms
    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        boolean res = false;
        for(int i = 0; i < m; i ++){
            for(int j = 0; j < n; j ++){
               if(dfs(board,word,i,j,0)){
                   res = true;
               }
            }
        }    
        return res;
    }     
    public boolean dfs(char[][] board, String word, int i, int j, int k){
        int m = board.length;
        int n = board[0].length;
        if(i < 0 || j < 0 || i >= m || j >= n){
            return false;
        }  
        if(board[i][j] == word.charAt(k)){
            char tmp = board[i][j];//记录当前位置
            board[i][j] = '#';//该位置已经被使用
            if(k == word.length() - 1){//匹配完毕
                return true;
            }else if(dfs(board, word, i - 1, j, k + 1)
            || dfs(board, word, i + 1, j, k + 1)
            || dfs(board, word, i, j - 1, k + 1)
            || dfs(board, word, i, j + 1, k + 1)){//遍历上下左右四个方向
                return true;
            }
            board[i][j] = tmp;//还原
        }     
        return false;
    }
}

class Solution {//14ms
    public boolean exist(char[][] board, String word) {
        for(int i = 0; i < board.length; i ++)
            for(int j = 0; j < board[0].length; j ++){
                if(dfs(board, word, i, j, 0))
                    return true;
            }
        return false;
    }
    private boolean dfs(char[][] board, String word, int i, int j, int k){
        if(k == word.length()) return true;
        if(i > board.length-1 || i <0 || j<0 || j >board[0].length - 1 || board[i][j] != word.charAt(k))
            return false;
        board[i][j] = '#';//改变该位置的值
        boolean flag =    dfs(board, word, i - 1, j, k + 1) ||
                            dfs(board, word, i, j - 1, k + 1) ||
                            dfs(board, word, i, j + 1, k + 1) ||
                            dfs(board, word, i + 1, j, k + 1);//遍历上下左右四个方向
        board[i][j] = word.charAt(k);//还原
        return flag;
    }
}

② 在discuss中看到，对处理已遍历的字符的处理方法不同。

class Solution {//9ms
    public boolean exist(char[][] board, String word) {
        int n = board.length, m = board[0].length;
        if (word.length() > n * m)  return false;
        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                if (dfs(board, word, i, j, 0))
                    return true;
            }
        }
        return false;
    }
    private boolean dfs(char[][] board, String word, int i, int j, int k){
        if (k == word.length()) return true;
        if (i < 0 || i == board.length || j < 0 || j == board[0].length)
            return false;
        if (board[i][j] != word.charAt(k))  return false;
        board[i][j] ^= 256;//改变该位置的值，将其变为非字母
        boolean flag = dfs(board, word, i - 1, j, k + 1) || dfs(board, word, i + 1, j, k + 1) 
                    || dfs(board, word, i, j - 1, k + 1) || dfs(board, word, i, j + 1, k + 1);
        board[i][j] ^= 256; //还原该位置的值
        return flag;
    }
}