Word Ladder II leetcode java
题目:
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
For example,
Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
Return
[
["hit","hot","dot","dog","cog"],
["hit","hot","lot","log","cog"]
]
Note:
- All words have the same length.
- All words contain only lowercase alphabetic characters.
题解:
答案是http://www.1point3acres.com/bbs/thread-51646-1-1.html 上面
iostreamin写的。
我就直接贴过来就好,这道题多读读代码看明白。
代码:
1 public ArrayList<ArrayList<String>> findLadders(String start, String end, HashSet<String> dict) {
2
3 HashMap<String, HashSet<String>> neighbours = new HashMap<String, HashSet<String>>();
4
5 dict.add(start);
6 dict.add(end);
7
8 // init adjacent graph
9 for(String str : dict){
10 calcNeighbours(neighbours, str, dict);
11 }
12
13 ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
14
15 // BFS search queue
16 LinkedList<Node> queue = new LinkedList<Node>();
17 queue.add(new Node(null, start, 1)); //the root has not parent and its level == 1
18
19 // BFS level
20 int previousLevel = 0;
21
22 // mark which nodes have been visited, to break infinite loop
23 HashMap<String, Integer> visited = new HashMap<String, Integer>();
24 while(!queue.isEmpty()){
25 Node n = queue.pollFirst();
26 if(end.equals(n.str)){
27 // fine one path, check its length, if longer than previous path it's valid
28 // otherwise all possible short path have been found, should stop
29 if(previousLevel == 0 || n.level == previousLevel){
30 previousLevel = n.level;
31 findPath(n, result);
32 }else {
33 // all path with length *previousLevel* have been found
34 break;
35 }
36 }else {
37 HashSet<String> set = neighbours.get(n.str);
38
39 if(set == null || set.isEmpty()) continue;
40 // note: I'm not using simple for(String s: set) here. This is to avoid hashset's
41 // current modification exception.
42 ArrayList<String> toRemove = new ArrayList<String>();
43 for (String s : set) {
44
45 // if s has been visited before at a smaller level, there is already a shorter
46 // path from start to s thus we should ignore s so as to break infinite loop; if
47 // on the same level, we still need to put it into queue.
48 if(visited.containsKey(s)){
49 Integer occurLevel = visited.get(s);
50 if(n.level+1 > occurLevel){
51 neighbours.get(s).remove(n.str);
52 toRemove.add(s);
53 continue;
54 }
55 }
56 visited.put(s, n.level+1);
57 queue.add(new Node(n, s, n.level + 1));
58 if(neighbours.containsKey(s))
59 neighbours.get(s).remove(n.str);
60 }
61 for(String s: toRemove){
62 set.remove(s);
63 }
64 }
65 }
66
67 return result;
68 }
69
70 public void findPath(Node n, ArrayList<ArrayList<String>> result){
71 ArrayList<String> path = new ArrayList<String>();
72 Node p = n;
73 while(p != null){
74 path.add(0, p.str);
75 p = p.parent;
76 }
77 result.add(path);
78 }
79
80 /*
81 * complexity: O(26*str.length*dict.size)=O(L*N)
82 */
83 void calcNeighbours(HashMap<String, HashSet<String>> neighbours, String str, HashSet<String> dict) {
84 int length = str.length();
85 char [] chars = str.toCharArray();
86 for (int i = 0; i < length; i++) {
87
88 char old = chars[i];
89 for (char c = 'a'; c <= 'z'; c++) {
90
91 if (c == old) continue;
92 chars[i] = c;
93 String newstr = new String(chars);
94
95 if (dict.contains(newstr)) {
96 HashSet<String> set = neighbours.get(str);
97 if (set != null) {
98 set.add(newstr);
99 } else {
HashSet<String> newset = new HashSet<String>();
newset.add(newstr);
neighbours.put(str, newset);
}
}
}
chars[i] = old;
}
}
private class Node {
public Node parent; //previous node
public String str;
public int level;
public Node(Node p, String s, int l){
parent = p;
str = s;
level = l;
}
}
Reference:http://www.1point3acres.com/bbs/thread-51646-1-1.html
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