127. Word Ladder

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Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord to endWord, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the word list

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["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.

虽然是两个字符串的转化问题，但DP貌似不好使，因为要判断是不是在set中；

然后想到的思路是构造图用最短路径，这样写下来的话先要map把set中的String映射到int，还要求出set里面可能的两两相邻的string构造一个二维连接矩阵，再用Dijkstra算法；这个貌似比较复杂，放弃

因为是最短路径的搜索，可以考虑DFS，BFS。这就比较粗暴，特别是DFS，而BFS就比较适用一点，因为只要求出最短的就好了，达到目标字符串就可以停止搜索，所以这个题目更适合用BFS。

bfs就是简单的dijkstra，就是步长为1时的特殊情况

可参考http://blog.csdn.net/zxzxy1988/article/details/8591890

public class Solution {    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {        int cnt = 2;        Set<String> reachSet = new HashSet<String>();        Set<String> nextReachSet = new HashSet<String>();        reachSet.add(beginWord);                while(!reachSet.isEmpty()) {        for(String word : reachSet) {        char[] cs = word.toCharArray();            for(int i=0; i<cs.length; i++) {            char orig = cs[i];            for(char c='a'; c<='z'; c++) {            if(c == orig)continue;            cs[i] = c;            String newBegin = new String(cs);                        if(newBegin.equals(endWord))return cnt;                        if(wordList.contains(newBegin)){            wordList.remove(newBegin);            nextReachSet.add(newBegin);            }                        // roll back            cs[i] = orig;            }            }        }                reachSet.clear();        reachSet = nextReachSet;        nextReachSet = new HashSet<String>();        cnt ++;        }            return 0;    }}

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