Leetcode|Word Ladder

最新推荐文章于 2020-05-19 11:34:31 发布

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本文介绍了一种使用广度优先搜索算法解决单词转换问题的方法。给定起始单词、目标单词及词典，寻找从起始单词到目标单词的最短转换序列，每次仅改变一个字母，且中间状态必须在词典中。

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, 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.

此题采用广度优先搜索，因为每个单词的变换只是遍历每一位，然后更换其他25个字母。如果新的单词在字典里面而且之前没有用过（用map记录），就加入进来。
map的value表示给单词是第几步。key值表示单词。

int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
        queue<string> cur;
        map<string,int> used;
        used[beginWord]=1;
        cur.push(beginWord);
        while(!cur.empty())
        {
            string word=cur.front();
            cur.pop();

            if(word==endWord) break;

            for(int i=0;i<word.size();i++){
                string tmp=word;
                for(char c='a';c<='z';c++){
                    if(c==word[i]) continue;
                    tmp[i]=c;
                    if(wordDict.count(tmp)!=0&&used.find(tmp)==used.end()){
                        used[tmp]=used[word]+1;
                        cur.push(tmp);
                    }
                }
            }
        }
        return used.find(endWord)==used.end()?0:used[endWord];
    }