Description

Given two words (beginWord and endWord), and a dictionary’s word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:

  1. Only one letter can be changed at a time
  2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
    Note:
  • Return an empty list if there is no such transformation sequence.
  • All words have the same length.
  • All words contain only lowercase alphabetic characters.
  • You may assume no duplicates in the word list.
  • You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:

beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output:

[
  ["hit","hot","dot","dog","cog"],
  ["hit","hot","lot","log","cog"]
]

Example 2:

Input:

beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output:

[]

Explanation:

The endWord "cog" is not in wordList, therefore no possible 
transformation.

分析

题目的意思是：给定一个开始单词，结束单词和一个单词词典，找出以开始单词为头，以结束单词为尾的所有最短路径。

  • 题目的目的是找出所有的路径，建立一个路径集paths，用以保存所有路径，然后是起始路径p，在p中先把起始单词放进去。然后定义两个整型变量level，和minLevel，其中level是记录循环中当前路径的长度，minLevel是记录最短路径的长度，这样的好处是，如果某条路径的长度超过了已有的最短路径的长度，那么舍弃，这样会提高运行速度，相当于一种剪枝。
  • 还要定义一个set变量words，用来记录已经循环过的路径中的词，然后就是BFS的核心了，循环路径集paths里的内容，取出队首路径，如果该路径长度大于level，说明字典中的有些词已经存入路径了，如果在路径中重复出现，则肯定不是最短路径，所以我们需要在字典中将这些词删去，然后将words清空，对循环对剪枝处理。然后我们取出当前路径的最后一个词，对每个字母进行替换并在字典中查找是否存在替换后的新词，这个过程在之前那道Word Ladder 词语阶梯里面也有。

代码

class Solution {
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
        vector<vector<string>> res;
        unordered_set<string> dict(wordList.begin(),wordList.end());
        vector<string> p{beginWord};
        queue<vector<string>> paths;
        paths.push(p);
        int level=1;
        int minLevel=INT_MAX;
        unordered_set<string> words;
        while(!paths.empty()){
            auto t=paths.front();
            paths.pop();
            if(t.size()>level){
                for(string w:words) dict.erase(w);
                words.clear();
                level=t.size();
                if(level>minLevel) break;
            }
            string last=t.back();
            for(int i=0;i<last.size();i++){
                string newLast=last;
                for(char ch='a';ch<='z';ch++){
                    newLast[i]=ch;
                    if(!dict.count(newLast)){
                        continue;
                    }
                    words.insert(newLast);
                    vector<string> nextPath=t;
                    nextPath.push_back(newLast);
                    if(newLast==endWord){
                        res.push_back(nextPath);
                        minLevel=level;
                    }else{
                        paths.push(nextPath);
                    }
                }
            }
            
        }
        return res;
    }
};

参考文献

​​[LeetCode] Word Ladder II 词语阶梯之二​​