解法

典型的深度优先遍历，注意isVisted的置1和置0

代码

class Solution {
    public boolean exist(char[][] board, String word) {

        int[][] isVisited = new int[board.length][board[0].length];
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if(f(board,isVisited,word,0,i,j)) return true;
            }
        }
        return false;
    }

    public boolean f(char[][] board, int[][] isVisited, String word, int index, int i, int j) {
        if (index==word.length()) return true;
        //超过了边界
        if (i > board.length-1||j > board[0].length-1||j<0||i<0) return false;
        //已经走过了
        if (isVisited[i][j] == 1) return false;
        if (board[i][j] == word.charAt(index)) {
            isVisited[i][j] = 1;//标志为已经走过
            //走下一步
            if (       !f(board, isVisited, word, index + 1, i, j + 1)
                    && !f(board, isVisited, word, index + 1, i, j - 1)
                    && !f(board, isVisited, word, index + 1, i + 1, j)
                    && !f(board, isVisited, word, index + 1, i - 1, j)){
                         isVisited[i][j]=0;
                return false;
            }
        } else return false;

        return true;
    }
}