79. 单词搜索

79. 单词搜索

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

 

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

 

 

class Solution {
public:
    int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    int vis[15][15];
    int n, m;
    bool dfs(vector<vector<char> >& board, string word, int x, int y, int k)
    {
        if(k == word.length())
            return true;
        for(int i = 0; i < 4; i++)
        {
            int tx = x + dir[i][0];
            int ty = y + dir[i][1];
            if(tx < 0 || ty < 0 || tx >=n || ty >= m || vis[tx][ty]) continue;
            if(board[tx][ty] == word[k])
            {
                vis[tx][ty] = 1;
                if(dfs(board, word, tx, ty, k + 1))
                    return true;
                vis[tx][ty] = 0;
            }
        }
        return false;
    }




    bool exist(vector<vector<char>>& board, string word) {
        memset(vis, 0, sizeof(vis));
        n = board.size();
        m = board[0].size();
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                vis[i][j] = 1;
                if(board[i][j] == word[0] && dfs(board, word, i, j, 1))
                    return true;
                vis[i][j] = 0;
            }
        }
        return false;



    }
};