Word Ladder

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

题意很容易看懂，所用的方法是广度优先搜索（BFS）算法。

int diffCount(const string& s1, const string& s2){int count = 0, len = s1.length();for (int i = 0; i < len; i++){if (s1[i] != s2[i])count++;}return count;}int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict){if (wordDict.size() == 0)return 0;queue<string> q;q.push(beginWord);unordered_map<string, int> visitLevel;//访问层次，每个字符串对应各自的最短访问长度visitLevel[beginWord] = 1;while (!q.empty()){string word = q.front(); q.pop();for (unordered_set<string>::iterator it = wordDict.begin(); it != wordDict.end(); ){string ladderWord = *it;if (diffCount(word, ladderWord) == 1)//word是否可以通过一次变更就达到下一个词典里有的某个单词{visitLevel[ladderWord] = visitLevel[word] + 1;//step+1if (*it == endWord)return visitLevel[ladderWord];q.push(ladderWord);//加入队列it = wordDict.erase(it);//从词典中删除访问过的单词}}}return 0;}