leetcode Word Ladder

LeetCode单词阶梯题解

最新推荐文章于 2018-06-12 10:14:22 发布

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最新推荐文章于 2018-06-12 10:14:22 发布

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原文链接：http://www.cnblogs.com/GadyPu/p/5034080.html

本文提供LeetCode单词阶梯问题的两种解决方案：一种是使用单向广度优先搜索的方法，另一种则是采用双向广度优先搜索的方式。通过这两种方法，文章详细介绍了如何找到从一个单词到另一个单词的最短转换路径。

题目连接

https://leetcode.com/problems/word-ladder/

Word Ladder

Description

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the word list
For example,

Given:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”]
As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”,
return its length 5.

class Solution {
public:
	int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {
		if (beginWord == endWord) return 0;
		unordered_set<string> vis;
		size_t n = endWord.size();
		typedef pair<string, int> PSI;
		queue<PSI> q;
		q.push(PSI(beginWord, 1));
		vis.insert(beginWord);
		while (!q.empty()) {
			PSI t = q.front(); q.pop();
			for (size_t i = 0; i < n; i++) {
				string x = t.first;
				for (int j = 0; j < 26; j++) {
					x[i] = 'a' + j;
					if (x == endWord) return t.second + 1;
					if (wordList.find(x) != wordList.end() && vis.find(x) == vis.end()) {
						q.push(PSI(x, t.second + 1));
						vis.insert(x);
					}
				}
			}
		}
		return 0;
	}
};

双向广搜

class Solution {
    using vec = vector<string>;
public:
    int ladderLength(string start, string end, unordered_set<string>& dic) {
        if(start == end) return 0;
        alpha = vector<bool>(26, false);
        for(auto &i: dic) { for(auto &j: i) alpha[j - 'a'] = true; }
        int cur = 0, rcur = 0;
        Q[cur].push_back(start);
        rQ[rcur].push_back(end);
        bfs.insert(start); rbfs.insert(end);
        auto expandState = [&, this](vec &from, vec &to, unordered_set<string> &bfs) {
            to.clear();
            for(auto &r: from) {
                for(int i = 0; i < (int)r.size(); i++) {
                    for(char j = 'a'; j <= 'z' ; j++) {
                        if(!alpha[j - 'a']) continue;
                        string temp = r;
                        temp[i] = j;
                        if(dic.find(temp) == dic.end() || bfs.find(temp) != bfs.end()) continue;
                        bfs.insert(temp);
                        to.push_back(temp);
                    }
                }
            }
        };
        for(int step = 1; step < 43 && !Q[cur].empty() && !rQ[rcur].empty(); step++) {
             if(Q[cur].size() <= rQ[rcur].size()) {
                expandState(Q[cur], Q[cur ^ 1], bfs);
                cur ^= 1;
                for(auto &r: Q[cur]) {
                    if(rbfs.find(r) != rbfs.end()) return step + 1;
                }
            } else {
                expandState(rQ[rcur], rQ[rcur ^ 1], rbfs);
                rcur ^= 1;
                for(auto &r: rQ[rcur]) {
                    if(bfs.find(r) != bfs.end()) return step + 1;
                }
            }
        }
        return 0;
    }
private:
    vec Q[2], rQ[2];
    vector<bool> alpha;
    unordered_set<string> bfs, rbfs;
};

转载于:https://www.cnblogs.com/GadyPu/p/5034080.html