[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:       }