Word Ladder问题及解法

问题描述：

Given two words (beginWord and endWord), and a dictionary's word list, 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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

示例：

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

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.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

问题分析：

对以此类问题，我们可以转换为一个图遍历的问题（BFS），图的每个节点间只有同一个位置上的字母不相同。设置visited标志数组，记录访问过的节点。wordList中包含了所有可以利用的节点。

过程详见代码：

class Solution {public:    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {unordered_set<string> visited;unordered_set<string> wordSet(wordList.begin(), wordList.end());int level = 1;int lastNum = 1;int curNum = 0;queue<string> que;que.push(beginWord);visited.insert(beginWord);while (!que.empty()){string cur = que.front();que.pop();lastNum--;for (int i = 0; i < cur.length(); i++){for (char c = 'a'; c <= 'z'; c++){if (c != cur[i]){string tcur = cur;tcur[i] = c;if (endWord == tcur && wordSet.find(tcur) != wordSet.end())return level + 1;if (wordSet.find(tcur) != wordSet.end() && visited.find(tcur) == visited.end()){curNum++;visited.insert(tcur);que.push(tcur);}}}}            if (lastNum == 0){lastNum = curNum;curNum = 0;level++;}}return 0;}};