[Leetcode 74] 92 Restore IP Addresses

Problem:

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)

 

Analysis:

Originally, I thought DFS may be a good solution. But since the brute-force enumeration is not that complicated, So I tried brute force method. The key observation is that:

1. to restore the IP address, we need to divide the original string into four sub-parts, thus need to choose 3 positions in the string to divide it.

2. for each substring, we need to judge whether it's valid or not. The rule is:

　　a. the value of the substring should in the range of [0, 255]

　　b. if it's not a 0, then no leading 0 is allowed. thus "010", "01", "00", "000" are invalid.

3. to get the final IP address, we need only concatinate as follows: s1 + "." + s2 + "." + s3 + "." + s4

 

For the dividing part, we can make use of some knowledge of IP address to reduce the work:

1. length of IP address should between 4 and 12 inclusively. Any given string's length less than 4 or greater than 12 can not be used as an IP address.

2. each sub-part's length should between 1 and 4 inclusively. Thus each for loop only need to check 3 division positions.

 

Code:

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> res;
        
        if (s.length() < 4 || s.length() > 12)
            return res;
            
        int len = s.length();
        
        for (int i=1; i<4 && i<len; i++)
            for (int j=i+1; j<i+4 && j<len; j++)
               for (int k=j+1; k<j+4 && k<len; k++) {
                       if (isValid(s, 0, i) && isValid(s, i, j) &&
                            isValid(s, j, k) && isValid(s, k, len)) {
                            string tmp = s.substr(0, i) + "." + 
                                            s.substr(i, j-i) + "." +
                                            s.substr(j, k-j) + "." + 
                                            s.substr(k, len-k);
                                
                            res.push_back(tmp);
                        }
                       
                   }
        
        return res;
    }
    
private:
    //  check sub-string s[s, e)'s validity
    bool isValid(string &s, int a, int e) {
        //only 1 bit, true anyway
        if (a+1 == e)
            return true;
        
        int val = 0;
        bool leading = false;
        for (int i=a; i<e; i++) {
            if (leading) {
                val = val * 10 + s[i] - '0';
            } else {
                if (s[i] == '0')
                    return false;
                else {
                    val = val * 10 + s[i]-'0';
                    leading = true;
                }
            }
        }
        
        if (val<=255 && val>=0)
            return true;
        else
            return false;
    }
};