word ladder

最新推荐文章于 2024-09-08 21:02:59 发布

原创最新推荐文章于 2024-09-08 21:02:59 发布 · 337 阅读

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#Leetcode #Graph #dfs

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本文探讨了如何在给定的单词列表中找到从起始单词到结束单词的最短转换路径，每一步只能改变一个字母且转换后的单词必须存在于列表中。通过使用广度优先搜索（BFS）算法实现这一目标。

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 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.

pro：一个start string，一个end string，一个dict，求将start变到end所需要的最小的步数，要求是每一步变换一个字符，中间结果都在dict中。

sol： 就是一个bfs，遍历每一位替换为其他字符所变换得到的单词，进队的条件是这个单词没被放进去过，这里用unordered_set,用map竟然tle......,以及单词在dict中。

code:

class Solution
{
public:
    int ladderLength(string start,string end,unordered_set<string> &dict)
    {
        int i,j,k,step;
        char ch;
        string temp,front;
        unordered_set<string> myset;
        myset.clear();
        queue< pair<string,int> > myque;
        while(!myque.empty())
            myque.pop();
        myque.push(make_pair(start,1));
        myset.insert(start);
        while(!myque.empty())
        {
            front = myque.front().first;
            step = myque.front().second;
            myque.pop();
            if(front==end) 
                return step;
            for(i=0;i<front.length();i++)
            {
                temp=front;
                for(j=0;j<26;j++)
                {
                    char ch=j+'a';
                    if(ch!=front[i])
                    {
                        temp[i]=ch;
                        if(temp==end) return step+1;
                        if(myset.find(temp)==myset.end()&&dict.find(temp)!=dict.end())
                        {
                            myset.insert(temp);
                            myque.push(make_pair(temp,step+1));
                        }
                    }
                }
            } 
        }
        return 0;
    }
};

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