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Monday, December 31, 2012

[LeetCode] Restore IP Addresses 解题报告

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)
» Solve this problem

[解题思路]
递归的将数字串分成四个部分,每个部分满足0<=p<=255。要注意一些边界case,比如010是没有意思的,“0.10.010.1”。

1:       vector<string> restoreIpAddresses(string s) {   
2:            vector<string> col;   
3:            string ip;   
4:            partitionIP(s, 0, 0, ip, col);   
5:            return col;   
6:       }   
7:       void partitionIP(string s, int startIndex, int partNum,    
8:       string resultIp, vector<string>& col)   
9:       {   
10:            //max: 3 bits per partition   
11:            if(s.size() - startIndex > (4-partNum)*3) return;   
12:            //min: 1 bit per partition   
13:            if(s.size() - startIndex < (4-partNum)) return;  
14:            if(startIndex == s.size() && partNum ==4)   
15:            {   
16:                 resultIp.resize(resultIp.size()-1);   
17:                 col.push_back(resultIp);   
18:                 return;   
19:            }   
20:            int num =0;   
21:            for(int i = startIndex; i< startIndex +3; i++)   
22:            {   
23:                 num = num*10 + (s[i]-'0');   
24:                 if(num<=255)   
25:                 {   
26:                      resultIp+=s[i];   
27:                      partitionIP(s, i+1, partNum+1, resultIp+'.', col);       
28:                 }    
29:                 if(num ==0)//0.0.0.0 valid, but need to avoid 0.1.010.01   
30:                 {   
31:                      break;   
32:                 }   
33:            }   
34:       }   



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