题目要求：

分析：
这道题目采用 DFS + 回溯 来做，它有四个方向来判断。设置一个visited数组，用来节点表示有没有访问过。
话说不清楚，直接上代码吧。

具体代码如下：

class Solution {
     public boolean exist(char[][] board, String word) {
         int m = board.length, n = board[0].length;
         boolean[][] visited = new boolean[m][n];
         for (int i = 0; i < m; i++) {
             for (int j = 0; j < n; j++) {
                 if (word.charAt(0) == board[i][j] && dfs(i, j, 0, word, board, visited))
                     return true;
             }
         }
         return false;
     }
    
    private boolean dfs(int i, int j, int index, String word, char[][] board, boolean[][] visited) {
        if(index == word.length())
            return true;
        int m = board.length, n = board[0].length;
        if(i < 0 || i >= m || j < 0 || j >= n || visited[i][j] || board[i][j] != word.charAt(index))
            return false;
        visited[i][j] = true;
        if(dfs(i + 1, j, index + 1, word, board, visited) || dfs(i, j + 1, index + 1, word, board, visited) 
          || dfs(i - 1, j, index + 1, word, board, visited) || dfs(i, j - 1, index + 1, word, board, visited))
            return true;
        visited[i][j] = false;
        return false;
    }
}