Leetcode-Resotre IP Addresses

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",

return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

Analysis:

This is a recursive problem. A string will be divided into four parts. For each part, we should determine whether it is a valid IP segment.

Solution:

 1 public class Solution {
 2     public List<String> restoreIpAddresses(String s) {
 3           List<String> res = new ArrayList<String>();
 4           if (s.length()==0) return res;
 5 
 6           res = restoreRecur(s,0,4);
 7           return res;        
 8     }
 9 
10     public List<String> restoreRecur(String s, int curIndex, int num){
11         List<String> res = new ArrayList<String>();
12         if (curIndex>=s.length()) return res;
13         
14         if (num==1){
15            String temp = s.substring(curIndex,s.length());
16            if (temp.length()>3) return res;
17            int val = Integer.parseInt(temp);
18            if (temp.length()==3 && val>=100 && val<=255){
19                res.add(temp);
20                return res;
21            }
22            if (temp.length()==2 && val>=10 && val<=99){
23                res.add(temp);
24                return res;
25            }
26            if (temp.length()==1){
27                res.add(temp);
28                return res;
29            }
30            return res;           
31         }
32         
33         if (curIndex+1>=s.length()) return res;
34         int end = curIndex+3;
35         if (curIndex+3>s.length())
36             end = s.length();
37         
38 
39         for (int i=curIndex+1;i<=end;i++){
40             String temp = s.substring(curIndex,i);
41             int val = Integer.parseInt(temp);
42             List<String> nextRes = new ArrayList<String>();
43             if (temp.length()==3 && val>=100 && val<=255){
44                 nextRes = restoreRecur(s,i,num-1);               
45             }
46             if (temp.length()==2 && val>=10 && val<=99){
47                 nextRes = restoreRecur(s,i,num-1);               
48             }
49             if (temp.length()==1){
50                 nextRes = restoreRecur(s,i,num-1);               
51             }
52             for (int j=0;j<nextRes.size();j++)
53                 res.add(temp+"."+nextRes.get(j));
54         }
55         return res;       
56         
57     } 
58 }

 

posted @ 2014-11-15 23:33  LiBlog  阅读(122)  评论(0)  编辑  收藏  举报