Leetcode 127 Word Ladder

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

1. Only one letter can be changed at a time
2. 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.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

每个单词每次只能变动一个字母变到字典中的其他单词，问从beginword变到endword最少需要多少步？

BFS，维护一个vis，表示已经到达过的单词，这样下次可以遇到可以不用搜索。

第一次提交超时了，以为BFS写搓了，调了半天发现是判断相邻单词的步骤超时。

原来版本的相邻词判断是这样的

    bool judge(string a,string b)
    {
        bool flag=false;
        for(int i=0;i::iterator it=wordList.begin();it!=wordList.end();it++)
        {
            unordered_set::iterator it2=it;
            for(it2++;it2!=wordList.end();it2++)
                if(judge(*it,*it2))
                {
                    mp[*it].push_back(*it2);
                    mp[*it2].push_back(*it);
                }
        }

怎么样？遍历所有单词组合判断他们是否相邻，复杂度n*n*len(word),

改成下面的做法过了，

对于每一个词，枚举每一位的变化为其他字母的情况，判断这个情况是否在集合中，复杂度n*len(word)*26*1，确实要快上一些，学到了！

class Solution {
public:
    int ladderLength(string beginWord, string endWord, unordered_set& wordList) {
        unordered_map> mp;
        wordList.insert(beginWord);
        wordList.insert(endWord);
        for(unordered_set::iterator it=wordList.begin();it!=wordList.end();it++)
        {
            string cc=*it;
            for(int i=0;i q,q2;
        unordered_map vis;
        q.push(beginWord);
        vis[beginWord]=true;
        int res=0,flag=1;
        while(!q.empty() || !q2.empty())
        {
            if(!q.empty())
            {
                string temp=q.front();
                q.pop();
                if(temp==endWord)
                {
                    res=flag;
                    break;
                }
                for(int i=0;i