[Leetcode] Word Search

最新推荐文章于 2025-03-02 14:59:52 发布

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本文介绍了一个算法问题，即在一个二维字符网格中查找指定单词是否存在。文章详细解释了使用深度优先搜索（DFS）策略来解决这一问题的方法，并给出了具体的C++实现代码。

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 =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

bool dfs(vector<vector<char> > &board, int x, int y, string word)  
    {  
        if(word.size() == 0)return true;  
  
        bool flag = false;  
        if(x-1>=0 && board[x-1][y] == word[0])  
        {  
            board[x-1][y] = '#';  
            flag = dfs(board, x-1, y, word.substr(1));  
            board[x-1][y] = word[0];  
        }  
          
        if(!flag && y-1>=0 && board[x][y-1] == word[0])  
        {  
            board[x][y-1] = '#';  
            flag = dfs(board, x, y-1, word.substr(1));  
            board[x][y-1] = word[0];  
        }  
  
        if(!flag && x+1<board.size() && board[x+1][y] == word[0])  
        {  
            board[x+1][y] = '#';  
            flag = dfs(board, x+1, y, word.substr(1));  
            board[x+1][y] = word[0];  
        }  
  
        if(!flag && y+1<board[0].size() && board[x][y+1] == word[0])  
        {  
            board[x][y+1] = '#';  
            flag = dfs(board, x, y+1, word.substr(1));  
            board[x][y+1] = word[0];  
        }  
          
        return flag;  
    }  
    bool exist(vector<vector<char> > &board, string word) {  
        // Note: The Solution object is instantiated only once.  
        if(word.size() < 1)return true;  
        int row = board.size();  
        int col = board[0].size();  
        set<string> st;  
        for(int i = 0; i < row; i++)  
            for(int j = 0; j < col; j++)  
                if(board[i][j] == word[0])  
                {  
                    board[i][j] = '#';  
                    if(dfs(board,i,j,word.substr(1)))  
                        return true;  
                    board[i][j] = word[0];  
                }  
        return false;  
    }