Leetcode: Word Search

最新推荐文章于 2019-07-08 21:27:00 发布

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#Leetcode #Word Search #dfs

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本文探讨如何在给定的二维字符网格中查找特定单词的路径，通过递归搜索相邻单元格，确保不重复使用相同的字母单元格。

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;
    }