1 class Solution {
 2 public:
 3     bool isIp(string s) {
 4         int len = s.size(), ip = 0;
 5         if (len == 0 || len > 4 || len > 1 && s[0] == '0') return false;
 6         for (int i = 0; i < len; i++) {
 7             ip = ip*10 + int(s[i] - '0');
 8         }
 9         if (ip > 255) return false;
10         return true;
11     }
12     
13     vector<string> restoreIpAddresses(string s) {
14         vector<string> result;
15         int len = s.size();
16         string ip1, ip2, ip3, ip4;
17         if (len < 4 || len > 12) return result;
18         for (int i = 0; i < 3; i++) {
19             if (i+1 > len) continue;
20             ip1 = s.substr(0, i+1);
21             if (isIp(ip1)) {
22                 for (int j = 0; j < 3; j++) {
23                     if (i+j+2 > len) continue;
24                     ip2 = s.substr(i+1, j+1);
25                     if (isIp(ip2)) {
26                         for (int k = 0; k < 3; k++) {
27                             if (i+j+k+3 > len) continue;
28                             ip3 = s.substr(i+j+2, k+1);
29                             ip4 = s.substr(i+j+k+3);
30                             if (isIp(ip3) && isIp(ip4)) {
31                                 result.push_back(ip1 + '.' + ip2 + '.' + ip3 + '.' + ip4);
32                             }
33                         }   
34                     }
35                 }
36             }
37         }
38         return result;
39     }
40 };

 

posted on 2015-03-23 05:41  keepshuatishuati  阅读(114)  评论(0编辑  收藏  举报