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)            return false;        if(word.length() == 0)            return true;                for(int i = 0; i< board.length; i++){            for(int j=0; j< board[0].length; j++){                if(board[i][j] == word.charAt(0)){                                        boolean rst = find(board, i, j, word, 0);                    if(rst)                        return true;                }            }        }        return false;    }        private boolean find(char[][] board, int i, int j, String word, int start){        if(start == word.length())            return true;                if (i < 0 || i>= board.length || j < 0 || j >= board[0].length || board[i][j] != word.charAt(start)){            return false; }                board[i][j] = '#'; // should remember to mark it        boolean rst = find(board, i-1, j, word, start+1) || find(board, i, j-1, word, start+1) ||                               find(board, i+1, j, word, start+1) || find(board, i, j+1, word, start+1));        board[i][j] = word.charAt(start);        return rst;    }

深搜，记得在每一次搜索的时候把开始元素mark成#,结束了以后再mark回来。