79. Word Search

##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
本题是典型的深度优先遍历，外层循环遍历整个二维数组，内层循环对每个节点上下左右做遍历
外层循环：

//外层循环
        for(int i=0;i<board.size();i++)
        {
            for(int j=0;j<board[0].size();j++)
            {
                if(search(board,word,0,i,j,used))
                    return true;
            }
        }

内层循环：

//DFS
        bool res = search(board,word,index+1,i-1,j,used)   
                || search(board,word,index+1,i+1,j,used)  
                || search(board,word,index+1,i,j-1,used)   
                || search(board,word,index+1,i,j+1,used);  

##代码

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(word.empty()||word.size()==0)
            return false;
        if(board.empty()||board.size()==0||board[0].size()==0)
            return false;
        vector<vector<bool>> used(board.size(),vector<bool>(board[0].size()));
        for(int i=0;i<board.size();i++)
        {
            for(int j=0;j<board[0].size();j++)
            {
                if(search(board,word,0,i,j,used))
                    return true;
            }
        }
        return false; 
    }
private:
    bool search(vector<vector<char>> board, string word, int index, int i, int j, vector<vector<bool>> used)
    {
        if(index==word.size())
            return true;
        if(i<0||j<0||i>=board.size()|| j>=board[0].size() || used[i][j] || board[i][j]!=word[index])
            return false; 
        used[i][j] = true;
        bool res = search(board,word,index+1,i-1,j,used)   
                || search(board,word,index+1,i+1,j,used)  
                || search(board,word,index+1,i,j-1,used)   
                || search(board,word,index+1,i,j+1,used);  
        used[i][j] = false;  
        return res; 
    }
};