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 }
LeetCode OJ - Word Ladder 2,布布扣,bubuko.com
时间： 2024-08-01 06:22:24