127. Word Ladder_127. word laddercsdn-优快云博客

本文链接：https://blog.youkuaiyun.com/gcola007/article/details/76651596

本文介绍了一种算法，用于寻找两个单词间最短的转换路径，每次仅改变一个字母且每一步都要形成有效的字典中的单词。通过使用队列和哈希集合实现双向广度优先搜索，有效地解决了这一问题。

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题目：

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the dictionary

For example,

Given:
start ="hit"
end ="cog"
dict =["hot","dot","dog","lot","log"]

As one shortest transformation is"hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length5.

两个for循环表示将start的每个字母从a遍历到z，同时与end,dict中的单词比较

外层的while循环是每当在dict中找到一个单词时，再从此单词开始

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        if(start.size()!=end.size())  return 0;
        if(dict.size()==0)  return 0;
        queue<string> queuetopush,queuetopop;
        queuetopop.push(start);
        int distance=1;
        while(dict.size()>0&&!queuetopop.empty()){
            while(!queuetopop.empty()){
                string str(queuetopop.front());
                queuetopop.pop();
                for(int i=0;i<str.size();i++){
                    for(char j='a';j<='z';j++){
                        if(str[i]==j)
                            continue;
                        char temp=str[i];
                        str[i]=j;
                        if(str==end)
                            return distance+1;
                        if(dict.count(str)){
                            queuetopush.push(str);
                            dict.erase(str);
                        }
                        str[i]=temp;
                    }
                }

            }
                distance++;
                swap(queuetopush,queuetopop);
        }
        return 0;
    }
};