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

刚拿到题目的时候主要是对题目理解了很久啊

其实意思很简单，每一个字符串中有一些数字和‘.’，‘.’的作用只是为了把数字分开，然后依次比较大小

在自己写的很土的方法中，就是简单的先把两个string分解到对应的vector当中，这里有几个需要注意的地方

两个字符串中的数字长度不是相等的，所以如果前面都相等还要看后面

11.22.33.00.00000 = 11.22.33

11.22.33.01 > 11.22.33

还有就是字符串最后面是没有‘.‘符号的，这里需要注意

class Solution {
public:
    int compareVersion(string version1, string version2) {
        vector<int> ver1;
	vector<int> ver2;
	int val1 = 0,val2 = 0;
	int i = 0,j = 0;
	while (i < version1.length())
	{
		if(version1[i] != '.')
		{
			val1 = val1*10+(version1[i]-'0');
		}
		else
		{
			ver1.push_back(val1);
			val1 = 0;
		}
		i++;
	}
	while (j < version2.length())
	{
		if (version2[j] != '.')
		{
			val2 = val2*10+(version2[j]-'0');
		}
		else
		{
			ver2.push_back(val2);
			val2 = 0;
		}
		j++;

	}
	ver1.push_back(val1);
	ver2.push_back(val2);
	bool flag = ver1.size() <= ver2.size();
	int length = flag? ver1.size():ver2.size();
	for (i = 0; i < length; i++)
	{
		if (ver1[i] > ver2[i])
		{
			return 1;
		}
		else if(ver1[i] < ver2[i])
		{
			 return -1;
		}
	}
	if (flag)
	{
		for (i = length; i < ver2.size(); i++)
		{
			if (ver2[i] != 0)
			{
				return -1;
			}
		}
		return 0;
	}
	else
	{
		for (i = length; i < ver1.size(); i++)
		{
			if (ver1[i] != 0)
			{
				return 1;
			}
		}
		return 0;
	}
    }
};

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int lev1=0,lev2=0;
        int id1=0,id2=0;
        while(id1!=version1.length()||id2!=version2.length()){
            lev1=0;
            while(id1<version1.length()){
                if(version1[id1]=='.'){
                    ++id1;
                    break;
                }
                lev1=lev1*10+(version1[id1]-'0');
                ++id1;
            }
            
            lev2=0;
            while(id2<version2.length()){
                if(version2[id2]=='.'){
                    ++id2;
                    break;
                }
                lev2=lev2*10+(version2[id2]-'0');
                ++id2;
            }
            
            if(lev1>lev2){
                return 1;
            }else if(lev1<lev2){
                return -1;
            }
        }
        return 0;//equal
    }
};

LeetCode—Compare Version Numbers