Compare two version numbers version1 and version1.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

class Solution 
{
public:
    int compareVersion(string version1, string version2) 
	{
        <span style="white-space:pre">	</span>vector<int> s1 = string_to_int(version1);
		vector<int> s2 = string_to_int(version2);
		int n = s1.size();
		int m = s2.size();
		int hold = n>m ? m:n;
		for(int i=0; i<hold; i++)
		{
			if(s1[i] > s2[i])
				return 1;
			else if(s1[i] < s2[i])
				return -1;
		}
		if(m==n)
			return 0;
		else if(m == hold)
		{
			for(int i=m; i<n; i++)
				if(s1[i] != 0)
					return 1;
			return 0;
		}
		else 
		{
			for(int i=n; i<m; i++)
				if(s2[i] != 0)
					return -1;
			return 0;
		}
    }
	vector<int> string_to_int(string s)
	{
		vector<int> res;
		int n = s.length();
		double temp = 0;
		bool falg = false;
		for(int i=0;i<n;i++)
		{
			if(s[i] == '.')
			{
				res.push_back(temp);
				falg = true;
				temp = 0;
				continue;
			}
			temp = temp*10 + s[i]-48; 
		}
		res.push_back(temp);
		return res;
	}
};

LeetCode--Compare Version Numbers