LeetCode

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:

//#define __MAIN__
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
using namespace std;

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        str = s.data();
        slen = strlen(str);
        res.clear();
        if(slen > 12){
            return res;
        }
        dfs(0, 0);
        
        return res;
    }
private:
    int slen;
    int a[4];
    vector<string> res;
    const char *str;
    char tmp[100];

    void dfs(int si, int ai) {
        int i;
        int num;
        string snum;
        
        if(ai == 4){
            if(si == slen){
                sprintf(tmp, "%d.%d.%d.%d", a[0], a[1], a[2], a[3]);
                res.push_back(string(tmp));
            }
            return;
        }
        
        if(si >= slen){
            return;
        }
        
        if(ai + (slen - si) / 3 > 4){
            return;
        }
        
        num = 0;
        snum = "";
        for(i = si; i < si + 3 && i < slen; ++i){
            if(snum.length() > 0 && snum[0] == '0'){
                //cannot add more digits after leading 0
                break;
            }
            num = num * 10 + (str[i] - '0');
            snum = snum + str[i];
            if(num > 255){
                break;
            }else{
                a[ai] = num;
                dfs(i + 1, ai + 1);
            }
        }
    }
};

#ifdef __MAIN__
#include <iostream>
int main()
{
    Solution sol;
    char str[100];
    vector<string> res;
    int i;

    while(gets(str) != NULL){
        res = sol.restoreIpAddresses(string(str));
        cout << "res_count = " << res.size() << endl;
        for(i = 0; i < res.size(); ++i){
            cout << res[i] << endl;
        }
        cout << endl;
    }

    return 0;
}
#endif