LeetCode - Restore IP Addresses

Restore IP Addresses

2013.12.31 18:06

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)

Solution:

　　Given a number string, find all of the possible IP address that it may represent in decimal form.

　　Finding all possible answer means we usually have to do a search, either DFS or BFS. This problem can be properly solved with DFS.

　　Time complexity is O(2^n), space compelxity is O(2^n) as well, where n is the length of the string.  Luckily the actual compelxity is much lower than 2^n, as each token in the ip address ranges in [0, 255], this natural limit helps us trim a lot of invalid search paths. By the way, the length n can be at most 12 (255.255.255.255), since the ip here only refers to IPv4, not including IPv6.

Accepted code:

 1 //#define __MAIN__

 2 #include <cstdio>

 3 #include <cstdlib>

 4 #include <cstring>

 5 #include <string>

 6 #include <vector>

 7 using namespace std;

 8 

 9 class Solution {

10 public:

11     vector<string> restoreIpAddresses(string s) {

12         // Note: The Solution object is instantiated only once and is reused by each test case.

13         str = s.data();

14         slen = strlen(str);

15         res.clear();

16         if(slen > 12){

17             return res;

18         }

19         dfs(0, 0);

20         

21         return res;

22     }

23 private:

24     int slen;

25     int a[4];

26     vector<string> res;

27     const char *str;

28     char tmp[100];

29 

30     void dfs(int si, int ai) {

31         int i;

32         int num;

33         string snum;

34         

35         if(ai == 4){

36             if(si == slen){

37                 sprintf(tmp, "%d.%d.%d.%d", a[0], a[1], a[2], a[3]);

38                 res.push_back(string(tmp));

39             }

40             return;

41         }

42         

43         if(si >= slen){

44             return;

45         }

46         

47         if(ai + (slen - si) / 3 > 4){

48             return;

49         }

50         

51         num = 0;

52         snum = "";

53         for(i = si; i < si + 3 && i < slen; ++i){

54             if(snum.length() > 0 && snum[0] == '0'){

55                 //cannot add more digits after leading 0

56                 break;

57             }

58             num = num * 10 + (str[i] - '0');

59             snum = snum + str[i];

60             if(num > 255){

61                 break;

62             }else{

63                 a[ai] = num;

64                 dfs(i + 1, ai + 1);

65             }

66         }

67     }

68 };

69 

70 #ifdef __MAIN__

71 #include <iostream>

72 int main()

73 {

74     Solution sol;

75     char str[100];

76     vector<string> res;

77     int i;

78 

79     while(gets(str) != NULL){

80         res = sol.restoreIpAddresses(string(str));

81         cout << "res_count = " << res.size() << endl;

82         for(i = 0; i < res.size(); ++i){

83             cout << res[i] << endl;

84         }

85         cout << endl;

86     }

87 

88     return 0;

89 }

90 #endif