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

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class Solution {
public:
    int compareVersion(string version1, string version2) {
        vector
  
    ver1;
	vector
   
     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
  
   lev2){
                return 1;
            }else if(lev1
   
    
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