Leetcode 79. Word Search

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.

    public boolean exist(char[][] board, String word) {        if (board == null || board.length == 0 || board[0].length == 0) return false;        int row = board.length, col = board[0].length;        for (int i = 0; i < row; i++) {            for (int j = 0; j < col; j++) {                if (word.charAt(0) != board[i][j]) continue;                if (exist(board, i, j, word, 0)) return true;            }        }        return false;    }    private boolean exist(char[][] board, int x, int y, String word, int start) {        if(start == word.length()) return true;        if(x < 0 || x >= board.length || y < 0 || y >= board[0].length || board[x][y] != word.charAt(start) || board[x][y] == '#') return false;        if (board[x][y] == word.charAt(start)) {            char c = board[x][y];            board[x][y] = '#';            boolean res = exist(board, x + 1, y, word, start + 1) ||                           exist(board, x - 1, y, word, start + 1) ||                          exist(board, x, y + 1, word, start + 1) ||                           exist(board, x, y - 1, word, start + 1);            board[x][y] = c;            return res;        }        return false;    }