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,
Givenboard=

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

分析

必然要对每一点的每一条路径进行深度遍历，遍历过程中一旦出现：

1.数组越界、2.该点已访问过、3.该点的字符和word对应的index字符不匹配

就要对该路径进行剪枝：

http://blog.csdn.net/happyaaaaaaaaaaa/article/details/50834335

http://www.cnblogs.com/yuzhangcmu/p/4040418.html

time 复杂度是m*n*4^(k-1). 也就是m*n*4^k.
m X n is board size, k is word size.

recuision最深是k层，recursive部分空间复杂度应该是O(k) + O(m*n)(visit array)

public class Solution {
    // recursion
    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;
    }
}