LeetCode Word Search

最新推荐文章于 2024-10-24 23:48:44 发布

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本文介绍了一种在二维网格中查找指定单词的算法。该算法通过递归地搜索相邻的字母块，确保不重复使用相同的字母单元。通过实例演示了如何在给定的网格中寻找目标单词。

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.

class Solution {
public:
	bool exist(vector<vector<char> > &board, string word) {
		if (word.size() == 0)
			return true;
		m = board.size();
		n = board[0].size();
		//上下左右扩展
		step[0][0] = -1;
		step[0][1] = 0;
		step[1][0] = 1;
		step[1][1] = 0;
		step[2][0] = 0;
		step[2][1] = -1;
		step[3][0] = 0;
		step[3][1] = 1;
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++)
				used[i][j] = false;
		}
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				if (board[i][j] == word[0]) {
					used[i][j] = true;
					flag = false;
					dfs(1, word.size(),i, j, board, word);
					if (flag)
						return true;
					used[i][j] = false;
				}
			}
		}
		return false;
	}
private:
	bool used[1000][1000];
	bool flag;
	int step[4][2];
	int m, n;

	void dfs(int depth, int maxDepth, int x, int y, vector<vector<char> > &board, string &word) {
	    if(flag)
	        return;
		if (depth == maxDepth) {
			flag = true;
			return;
		}
		for (int i = 0; i < 4; i++) {
			int nx = x + step[i][0];
			int ny = y + step[i][1];
			if (0 <= nx && nx < m && 0 <= ny && ny < n && !used[nx][ny] && board[nx][ny] == word[depth]) {
				used[nx][ny] = true;
				dfs(depth + 1, maxDepth, nx, ny, board, word);
				used[nx][ny] = false;
			}
		}
	}
};