【Leetcode】Word Ladder

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;    }};