leetcode 79:Word Search（redo）

题目：

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 .

思路：

用DFS来做，对某一个元素的上下左右来进行递归判断是否满足。

实现如下：

 class Solution {
    public:
         bool exist(vector<vector<char> > &board, string word) {
             m=board.size();
             n=board[0].size();
            for(int x=0;x<m;x++)
                for(int y=0;y<n;y++)
                {
                    if(isFound(board,word.c_str(),x,y))
                        return true;
                }
            return false;
        }
    private:
        int m;
        int n;
        bool isFound(vector<vector<char> > &board, const char* w, int x, int y)
        {
            if(x<0||y<0||x>=m||y>=n||board[x][y]=='\0'||*w!=board[x][y])
                return false;
            if(*(w+1)=='\0')
                return true;
            char t=board[x][y];
            board[x][y]='\0';
            if(isFound(board,w+1,x-1,y)||isFound(board,w+1,x+1,y)||isFound(board,w+1,x,y-1)||isFound(board,w+1,x,y+1))
                return true; 
            board[x][y]=t;
            return false;
        }
    };