题目：

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

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


解题思路：采用BFS搜索。每个节点采用pair表示，每个pair的第一个元素是字符串本身，第二个元素是所在层次。


代码：

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        queue<pair<string,int>> WordCandidate;
        if(start.empty()||end.empty())return 0;
        int size=start.size();
        if(start==end)return 1;
        WordCandidate.push(make_pair(start,1));
        while(!WordCandidate.empty()){
            pair<string,int> CurrWord(WordCandidate.front());
            WordCandidate.pop();
            for(int i=0;i<size;i++){
                for(char c='a';c<='z';c++){
                    swap(c,CurrWord.first[i]);
                    if(CurrWord.first==end)return CurrWord.second+1;
                    if(dict.count(CurrWord.first)>0){
                        WordCandidate.push(make_pair(CurrWord.first,CurrWord.second+1));
                        dict.erase(CurrWord.first);
                    }
                    swap(c,CurrWord.first[i]);
                }
            }
        }
    return 0;
    }
};