[Leetcode] 127. Word Ladder 解题报告_leetcode 127 题-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/71078120

本文介绍了一种利用广度优先搜索算法寻找两个单词间最短转换路径的方法。通过从两端同时进行搜索，并选择较小的搜索空间来减少计算量。

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题目：

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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

思路：

为了减少搜索量，我们采用从两头向中间同时搜索的策略。也就是说，分别设立两个hash表，一个初始化为beginWord，一个初始化为endWord，然后在wordList里面查找和hash表相邻的节点。一旦发现了，则有两种情况：1）该相邻节点已经存在于另外一个hash表中。此时说明从两头开始的搜索已经相遇，我们找到了最短路径，所以直接返回；否则将该相邻节点插入hash表中，继续查找即可。

本质上来讲，本题也算是很标准的广度优先搜索了。为了减少搜索量，我们每次搜索完一层后，都从size比较小的一个hash表开始下一层的搜索。当然这只是属于实现层面的小技巧而已。

代码：

class Solution {
public:
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {
        int res = 1;
        unordered_set<string> set1{beginWord};
        unordered_set<string> set2{endWord};
        while(set1.size() > 0) {
            res++;
            unordered_set<string> set;
            for(auto word : set1) {
                wordList.erase(word);
            }
            for(auto word : set1) {
                for(size_t i = 0; i < word.size(); ++i) {
                    string next = word;
                    for(char c = 'a'; c != 'z'; ++c) {
                        next[i] = c;
                        if(wordList.find(next) == wordList.end()) {
                            continue;
                        }
                        if(set2.find(next) != set2.end()) {
                            return res;
                        }
                        set.insert(next);
                    }
                }
            }
            set1 = set.size() < set2.size() ? set : set2;   // set1 is the one with smaller size
            set2 = set.size() < set2.size() ? set2 : set;   // set2 is the one with larger size
        }
        return 0;
    }
};