lintcode - 恢复ip地址 

 1 class Solution {
 2 public:
 3     /*
 4      * @param s: the IP string
 5      * @return: All possible valid IP addresses
 6      */
 7     vector<string> restoreIpAddresses(string &s) {
 8         // write your code here
 9         vector<string> ans;
10         dfs(s, 0, ans, "", 0);
11         return ans;
12     }
13     void dfs(string &s, int pos, vector<string> &ans, string res, int cnt){
14         if(pos >= s.length() && cnt == 4){
15                 ans.push_back(res);
16                 return ;
17         } else {
18                 for(int i = 0; i <= 2; ++i){
19                         if(pos + i < s.length()){
20                                 string tmp = s.substr(pos, i + 1);\
21                                 int num = atoi(tmp.c_str());
22                                 if((num == 0 && tmp.length() != 1) || (num != 0 && tmp[0] == '0')) continue;
23                                 string nres = "";
24                                 if(i == 2){
25                                        if(num < 256){
26                                         if(pos != 0)
27                                                 nres = res + "." + tmp;
28                                         else    nres = res + tmp;
29                                         dfs(s, pos + i  + 1, ans, nres, cnt + 1);
30                                        }
31                                 } else {
32                                         if(pos != 0)
33                                                 nres = res + "." + tmp;
34                                                 else nres = res + tmp;
35                                         dfs(s, pos + i + 1, ans, nres, cnt + 1);
36                                 }
37                         }
38                 }
39         }
40     }
41 };

ip地址的每个区域值不大于255 且不能有前导零

posted @ 2017-09-18 20:11  GeniusYang  阅读(238)  评论(0)  编辑  收藏  举报