[LeetCode] 139: Word Search

最新推荐文章于 2022-01-28 17:24:00 发布

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本文介绍了一个二维网格中查找指定单词的算法实现。采用深度优先搜索（DFS）策略，从每个可能的起点开始遍历，检查是否能按顺序匹配网格中的字母形成目标单词。

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[Problem]

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.

[Solution]

class Solution {
public:
	/**
	 * DFS
	 */
	bool DFS(vector<vector<char> > &board, bool **visited, int x, int y, string word, int begin){
		// empty string
		if(begin == word.size()){
			return true;
		}
		// end of DFS
		if(begin == word.size()-1 && board[x][y] == word[begin]){
			return true;
		}

		// DFS
		visited[x][y] = true;
		int table[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
		for(int i = 0; i < 4; ++i){
			if(x+table[i][0] < 0 || x+table[i][0] == board.size() || y+table[i][1] < 0 || y+table[i][1] == board[0].size() || visited[x+table[i][0]][y+table[i][1]] || board[x+table[i][0]][y+table[i][1]] != word[begin+1])continue;

			// visit
			visited[x+table[i][0]][y+table[i][1]] = true;
			if(DFS(board, visited, x+table[i][0], y+table[i][1], word, begin+1))return true;
			visited[x+table[i][0]][y+table[i][1]] = false;
		}
		return false;
	}
	/**
	 * if word exist in board
	 */
	bool exist(vector<vector<char> > &board, string word) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		// empty string[x+table[i][0]][y+table[i][1]]
		if(word.size() == 0)return true;

		// invalid
		if(board.size() == 0 || board[0].size() == 0 || board.size()*board[0].size() < word.size())return false;

		// initial
		bool **visited = new bool*[board.size()];
		for(int i = 0; i < board.size(); ++i){
			visited[i] = new bool[board[i].size()];
		}

		// search
		for(int x = 0; x < board.size(); ++x){
			for(int y = 0; y < board[x].size(); ++y){
				if(board[x][y] == word[0]){
					// initial visited
					for(int i = 0; i < board.size(); ++i){
						for(int j = 0; j < board[i].size(); ++j){
							visited[i][j] = false;
						}
					}
					// DFS
					if(DFS(board, visited, x, y, word, 0)){
						return true;
					}
				}
			}
		}
		return false;
	}
};