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.

class Solution {  
private:  
    bool dfs(vector<vector<char>>&board, string word, int count, int row, int col, vector<vector<int>>&visited)  
    {  
        if (count == word.size() - 1)  
            return true;  
        visited[row][col] = 1;//已经访问  
        if (row + 1 < board.size() &&visited[row+1][col]==0&& board[row + 1][col] == word[count + 1] )  
            if (dfs(board, word, count + 1, row + 1, col, visited))  
                return true;  
        if (row - 1 >= 0 &&visited[row-1][col]==0&& board[row - 1][col] == word[count + 1] )  
            if (dfs(board, word, count + 1, row - 1, col, visited))  
                return true;  
        if (col + 1 < board[0].size() &&visited[row][col+1]==0&& board[row][col + 1] == word[count + 1] )  
            if (dfs(board, word, count + 1, row, col + 1, visited))  
                return true;  
       if (col - 1 >= 0 &&visited[row][col-1]==0&& board[row][col - 1] == word[count + 1] )  
            if (dfs(board, word, count + 1, row, col - 1, visited))  
                return true;  
        visited[row][col] = 0;//失败之后要恢复到未访问的状态  
        return false;  
    }  
public:  
    bool exist(vector<vector<char>>& board, string word) {  
        if (word.size() == 0)  
            return true;  
        int row = board.size();  
        int col = board[0].size();  
        vector<vector<int>>visited(row, vector<int>(col));  
        //0----unvisited  
        //1----visited  
        if (row == 0 || col == 0)  
            return false;  
        for (int i = 0; i < row; i++)  
        {  
           for (int j = 0; j < col; j++)  
            {  
                if (board[i][j] == word[0])  
                {  
                    if (dfs(board, word, 0, i, j, visited))//开始递归  
                        return true;  
                }  
            }  
        }  
        return false;  
    }  
};