LeetCode-Word Ladder

LiBlog發表於2015-01-05
LeetCode

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

  1. Only one letter can be changed at a time
  2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["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.

Solution (Double direct BFS):

 1 public class Solution {
 2     public int ladderLength(String start, String end, Set<String> dict) {
 3         //Double direct BFS
 4         List<String> list1 = new ArrayList<String>();
 5         Map<String,Integer> set1 = new HashMap<String,Integer>();
 6         List<String> list2 = new ArrayList<String>();
 7         Map<String,Integer> set2 = new HashMap<String,Integer>();
 8         int index1 = 0;
 9         int index2 = 0;
10         list1.add(start);
11         list2.add(end);
12         set1.put(start,1);
13         set2.put(end,1);
14         String curStr1, curStr2;
15         String interStr="";
16         while (true){
17             boolean find = false;
18             
19             //No solution
20             if (index1>=list1.size())
21                return 0;
22             
23             curStr1 = list1.get(index1);
24             for (int i=0;i<curStr1.length();i++){
25                 char array[] = curStr1.toCharArray();
26                 for (array[i]='a';array[i]<='z';array[i]++){
27                     String temp = new String(array);
28                     if (set1.containsKey(temp)) continue;
29                     if (dict.contains(temp)){
30                         list1.add(temp);
31                         set1.put(temp,set1.get(curStr1)+1);
32                         
33                         if (set2.containsKey(temp)){
34                             interStr = temp;
35                             find = true;
36                             break;
37                         }
38                     }
39                 }
40             }
41             index1++;
42             if (find)
43                 break;
44             
45             if (index2>=list2.size())
46                 return 0;
47                 
48             curStr2 = list2.get(index2);
49             for (int i=0;i<curStr2.length();i++){
50                 char array[] = curStr2.toCharArray();
51                 for (array[i]='a';array[i]<='z';array[i]++){
52                     String temp = new String(array);
53                     if (set2.containsKey(temp)) continue;
54                     if (dict.contains(temp)){
55                         list2.add(temp);
56                         set2.put(temp,set2.get(curStr2)+1);
57                         
58                         if (set1.containsKey(temp)){
59                             interStr = temp;
60                             find = true;
61                             break;
62                         }
63                     }
64                 }
65             }
66            
67             
68             index2++;
69             if (find)
70                 break;
71         }
72         
73         
74         //Add path from pos direct
75         int dis = 0;
76         dis += set1.get(interStr);
77         dis += set2.get(interStr);
78         dis--;
79         
80         return dis;
81     }
82     
83     
84    
85 }

 

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