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.

大致题意：比较版本号，比较的两个元素由顿号分割的字符串组成

int compareVersion(string version1, string version2) {    version1+=‘.‘;    version2+=‘.‘;    int l1=version1.length(),l2=version2.length(),i,j,num1,num2;    i=0;    j=0;    while (i<l1 || j<l2)    {        num1=0;        while (i<l1 && version1[i]!=‘.‘)        {            num1=num1*10+version1[i]-‘0‘;            ++i;        }        if (version1[i]==‘.‘) ++i;        num2=0;        while (j<l2 && version2[j]!=‘.‘)        {            num2=num2*10+version2[j]-‘0‘;            ++j;        }        if (version2[j]==‘.‘) ++j;        if (num1!=num2)            return num1>num2?1:-1;    }    return 0;  }

这个题有意思的在于统一化的处理

先在串尾加顿号，这样每次都可以以顿号为判定条件

其次while (i<l1 || j<l2)这一句也很有意思，我开始写的是 while(i<l1 && j<l2),这样的话每次对串长不同的要在while后判定处理

if (i==l1 && j!=l2) return -1;    else if (i!=l1 && j==l2) return 1;

因为前面都相同，没完的串肯定版本新，于是光荣的WA了，有数据是1 1.0，23333

所以在一个串结束后，未完串是否为0，也是要看的，这里仅需要把循环条件修改就行了

leetcode:Compare Version Numbers