127. Word Ladder-未完待续

最初的思路（TLE，最大数据集wordList约2800个词）：
维护两个集合，分别代表beginWord的n级关系和endWord的n级关系；
若beginSet和endSet在第n级关系的交集为空，则beginSet和endSet分别去wordList中扩展下一级关系。
若beginSet和endSet在第n级关系的交集不为空，则返回关系级数。
代码如下：

public class Solution {    public int countDiffCh(String word1, String word2){        int count = 0;        for(int i = 0; i < word1.length(); i ++){            if (word1.charAt(i) != word2.charAt(i))                count ++;        }        return count;    }    //add the next level neighbor into set, and remove them from the wordList    //return true: changed, false otherwise;    public boolean addNewNeighbor(Set<String> set, LinkedList<String> wordList){        boolean flag = false;        Set<String> newSet = new HashSet<String>();        for(String wordInSet : set){            int p = 0;            while(p < wordList.size()){                if (countDiffCh(wordInSet, wordList.get(p)) == 1){                    newSet.add(wordList.get(p));                    wordList.remove(p);                    flag = true;                }else{                    p ++;                }            }            if (wordList.size() == 0)                break;        }        set.addAll(newSet);        return flag;    }    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {        Set<String> beginSet = new HashSet<String>();        beginSet.add(beginWord);        Set<String> endSet = new HashSet<String>();        endSet.add(endWord);        LinkedList<String> beginList = new LinkedList<String>();//lengths of words in the wordList should be all the same        LinkedList<String> endList = new LinkedList<String>();        for (String word : wordList){            if (! word.equals(beginWord))                beginList.add(word);            if (! word.equals(endWord))                endList.add(word);        }        int pathLen = 1;        Set<String> tmp;        while(true){            tmp = new HashSet(beginSet);            tmp.retainAll(endSet);            if (tmp.size() != 0){                break;            }            if ((beginList.size() == 0 && endList.size() == 0) ||                 (addNewNeighbor(beginSet, beginList) == false && addNewNeighbor(endSet, endList) == false)){                pathLen = 0;                break;            }            pathLen ++;        }        return pathLen;    }}

解决办法：
把wordList中的词和Set中的词一一比较太费时，可以用wordList中的词去产生所有相差一个字母的情况，然后判断Set.contains(word)，不超时
https://discuss.leetcode.com/topic/64729/share-my-easy-bfs-solution-java
这个代码实现更简洁：https://discuss.leetcode.com/topic/17890/another-accepted-java-solution-bfs/2

改写如下，大数据还是TLE：

public class Solution {    //add the next level neighbor into set, and remove them from the wordList    //return true: changed, false otherwise;    public boolean addNewNeighbor(Set<String> set, LinkedList<String> wordList){        boolean flagNewNeighbor = false;        int p = 0;        boolean flag = false; // cur word is a new neighbor;        Set<String> newSet = new HashSet<String>();        while(p < wordList.size()){            flag = false;            for (int i = 0; i < wordList.get(p).length(); i ++){                char[] chs = wordList.get(p).toCharArray();                for (char ch = 'a'; ch <= 'z'; ch ++){                    if (ch == wordList.get(p).charAt(i))                        continue;                    else                        chs[i] = ch;                    String newWord = new String(chs);                    if (set.contains(newWord)){                        newSet.add(wordList.get(p));                        wordList.remove(p);                        flag = true;                        break;                    }                }                if (flag == true)                    break;            }            if (flag == false)                p ++;            else                flagNewNeighbor = true;        }        set.addAll(newSet);        return flagNewNeighbor;    }    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {        Set<String> beginSet = new HashSet<String>();        beginSet.add(beginWord);        LinkedList<String> beginList = new LinkedList<String>();//lengths of words in the wordList should be all the same        for (String word : wordList){            if (! word.equals(beginWord))                beginList.add(word);        }        int pathLen = 1;        while(true){            if (beginSet.contains(endWord)){                break;            }            if (beginList.size() == 0  || addNewNeighbor(beginSet, beginList) == false){                pathLen = 0;                break;            }            pathLen ++;        }        return pathLen;    }}