我发现在leetcode上做题，当我出现TLE问题时，往往是代码有漏洞，有些条件没有考虑到，这道题又验证了我这一想法。
这道题是在上一道的基础上进一步把所有可能得转换序列给出。
同样的先是BFS，与此同时需要一个hashMap记录下每个节点，和他所有父节点的对应关系，然后通过DFS，回溯所有可能的路径。
下面是AC代码。
1 /** 2 * Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, 3 * @param start 4 * @param end 5 * @param dict 6 * @return 7 */ 8 public ArrayList<ArrayList<String>> findLadders(String start, String end, HashSet<String> dict) { 9 10 //for BFS 11 LinkedList<String> queue = new LinkedList<String>(); 12 //store the words that have visited and in the dict, and its corresponding level 13 HashMap<String,Integer> visited = new HashMap<String,Integer>(); 14 //the level information for every word 15 LinkedList<Integer> level = new LinkedList<Integer>(); 16 //the word and its parents 17 HashMap<String,ArrayList<String>> wP = new HashMap<String,ArrayList<String>>(); 18 queue.offer(start); 19 level.offer(1); 20 wP.put(queue.peek(), null);//start has no parents; 21 visited.put(start, 1); 22 while(!queue.isEmpty() && (!visited.containsKey(end) || level.peek() == visited.get(end)-1)){ 23 String par = queue.poll(); 24 int le = level.poll(); 25 26 //for every character in the word 27 for(int i=0;i<par.length();i++){ 28 29 char[] words = par.toCharArray(); 30 char o = words[i];//the original word 31 for(char c='a';c<='z';c++) 32 { 33 if(c!=o){ 34 //subsitude by another char 35 words[i] = c; 36 String changed = new String(words); 37 //the last-1 level . 38 if(changed.equals(end)) 39 { 40 visited.put(changed, le+1);// it.s very important!!!! Dont't forget!! 41 42 if(wP.containsKey(end)){ 43 ArrayList<String> p = wP.get(end); 44 p.add(par); 45 wP.put(end, p); 46 }else{ 47 ArrayList<String> p = new ArrayList<String>(); 48 p.add(par); 49 wP.put(end, p); 50 } 51 } 52 //return le+1; 53 else if((visited.get(changed)==null || visited.get(changed) == le+1) && dict.contains(changed)){ 54 //the condition is very important!!! otherwise, there will be duplicate. 55 if(!visited.containsKey(changed)) 56 { 57 queue.offer(changed); 58 level.offer(le+1); 59 } 60 visited.put(changed,le+1); 61 62 //update the word and his parents information 63 if(wP.containsKey(changed)){ 64 ArrayList<String> p = wP.get(changed); 65 p.add(par); 66 wP.put(changed, p); 67 }else{ 68 ArrayList<String> p = new ArrayList<String>(); 69 p.add(par); 70 wP.put(changed, p); 71 } 72 } 73 } 74 } 75 76 } 77 } 78 ArrayList<ArrayList<String>> fl =new ArrayList<ArrayList<String>>(); 79 //it's very important!!! to Check whether it has such path 80 if(!wP.containsKey(end)) 81 return fl; 82 traceback(wP,end,fl, null); 83 84 return fl; 85 } 86 /** 87 * DFS ,对每个节点的父节点进行深度遍历 88 * @param wP 89 * @param word 90 * @param fl 91 * @param cur 92 */ 93 private void traceback(HashMap<String,ArrayList<String>> wP, String word, ArrayList<ArrayList<String>> fl, 94 ArrayList<String> cur){ 95 if(wP.get(word)==null) 96 { 97 ArrayList<String> next = new ArrayList<String>(); 98 next.add(word); 99 if(cur!=null && cur.size()>0) 100 next.addAll(cur); 101 fl.add(next); 102 return; 103 } 104 for(String p: wP.get(word)){ 105 ArrayList<String> next = new ArrayList<String>(); 106 next.add(word); 107 if(cur!=null && cur.size()>0) 108 next.addAll(cur); 109 traceback(wP, p, fl, next); 110 } 111 }