79. 单词搜索-c++

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true
示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true
示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int x=board.size();
        int y=board[0].size();
        for(int i=0;i<x;i++)
        {
            for(int j=0;j<y;j++)
            {
                if(backtrack(board,word,i,j,0))
                {
                    return true;
                }
            }
        }
        return false;
    }
    bool backtrack(vector<vector<char>>& board, string word,int i,int j,int k)
    {
        if(k==word.size())
        {
            return true;
        }
        if(i<0||j<0||i>=board.size()||j>=board[0].size()||board[i][j]!=word[k])
        {
            return false;
        }
        char a=board[i][j];
        board[i][j]='\0';
        bool re=(backtrack(board,word,i-1,j,k+1)||
        backtrack(board,word,i+1,j,k+1)||
        backtrack(board,word,i,j+1,k+1)||
        backtrack(board,word,i,j-1,k+1));
 
        board[i][j]=a;
        return re;
    }
};