题解

1. dfs ，对棋盘里的每个点都dfs一遍，看是否有合适的字符串。
2. 当找到最后一个字符位置，并且最后一个字符，并且当前字符串匹配，那么为真。
3. 注意回溯之后的标记需要改成false , 因为需要回溯进行查找。

class Solution {

    public static boolean vis[][] = new boolean[8][8];
    public static int dir[] = {-1,0,1,0,-1};
    public static boolean ans;

    public static void dfs(char[][] board, String word, int x, int y, int n ){
        // System.out.println("n:  " + n);

        if(n == word.length() -1 && board[x][y] == word.charAt(n) ){
            ans = true;
            return;
        }

        for(int i = 0; i < 4; i++){
            int dx = x + dir[i];
            int dy = y + dir[i+1];

            // System.out.println(dx + " " + dy);
            
            if(dx >= 0 && dx < board.length && dy >=0 && dy < board[0].length && !vis[dx][dy] && board[x][y] == word.charAt(n)) {
                vis[dx][dy] = true;

                // System.out.println(word.charAt(n) + " " + n);
                // System.out.println(x + " " + y);
                // System.out.println("d " + dx + " " + dy);

                dfs(board, word, dx, dy, (n + 1));
                vis[dx][dy] = false;
            }

        }

    }
    public boolean exist(char[][] board, String word) {
        ans = false;


        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(word.length() == 1 && board[i][j] == word.charAt(0)){
                    ans = true;
                }
                else if(board[i][j] == word.charAt(0)) {
                    vis[i][j] = true;
                    dfs(board, word, i, j, 0);
 
                    vis[i][j] = false;
                }
                
                
            }
        }
        // System.out.println(ans);
        
        return ans;
//        System.out.println();


    }
}