Leetcode 165. Compare Version Numbers

Compare two version numbers version1 and version2.
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

s思路：
1. 由于version长度不规则，所以需要用while取数据。 取版本号时，由于版本可能超过int的范围，需要不转还直接比较吗？
2. bug:在写的时候，刚开始判断两个string只要有一个遍历完就退出循环，即：while(i

class Solution {
public:
    int compareVersion(string version1, string version2) {
        //
        version1.push_back('.');
        version2.push_back('.');     
        int i=0,j=0,m1=version1.size(),m2=version2.size();
        //while(i<m1&&j<m2){//不专业的写法
        while(i<m1||j<m2){
            int num1=0;
            if(i<m1&&version1[i]!='.'){
                while(i<m1&&version1[i]!='.'){//加保护
                    num1=num1*10+version1[i]-'0';
                    i++;
                }    
            }else
                i++;
            int num2=0;
            if(j<m2&&version2[j]!='.'){
                while(j<m2&&version2[j]!='.'){//加保护
                    num2=num2*10+version2[j]-'0';
                    j++;
                }    
            }else
                j++;
            //cout<<num1<<endl<<num2<<endl;
            if(num1>num2) return 1;
            if(num1<num2) return -1;
        }   
        return 0;
    }
};