首页 > 程序开发 > 软件开发 > 其他 >

Leetcode 127 Word Ladder

2016-10-27

Given two words (beginWordandendWord), and a dictionary s word list, find the length of shortest transformation sequence frombeginWordtoendWord, such that: Only one letter can be changed at a time Each intermediate word must exist in the word list

Given two words (beginWordandendWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWordtoendWord, 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 length5.

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<a.size();i++)  
    {  
        if(a[i]!=b[i])  
        {  
            if(flag) return false;  
            flag=true;  
        }  
    }  
    return flag;  
}  
    for(unordered_set<string>::iterator it=wordList.begin();it!=wordList.end();it++)  
    {  
        unordered_set<string>::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<string>& wordList) {  
        unordered_map<string,vector<string>> mp;  
        wordList.insert(beginWord);  
        wordList.insert(endWord);  
        for(unordered_set<string>::iterator it=wordList.begin();it!=wordList.end();it++)  
        {  
            string cc=*it;  
            for(int i=0;i<cc.size();i++)  
            {  
                string c=*it;  
                for(char j=&#39;a&#39;;j<=&#39;z&#39;;j++)  
                {  
                    c[i]=j;  
                    if(wordList.find(c)!=wordList.end())  
                    {  
                        mp[*it].push_back(c);  
                    }  
                }  
            }  
        }  
        queue<string> q,q2;  
        unordered_map<string,bool> 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<mp[temp].size();i++)  
                {  
                    string tt=mp[temp][i];  
                    if(vis[tt]) continue;  
                    vis[tt]=true;  
                    q2.push(tt);  
                }  
            }  
            else  
            {  
                flag++;  
                swap(q,q2);  
            }  
        }  
          
        return res;  
    }  
};
相关文章
最新文章
热点推荐