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:
代码语言：javascript复制0.1 < 1.1 < 1.2 < 13.37版本号比较问题。
模拟，练练手，注意边界情况，点可能有多个
代码语言：javascript复制class Solution {
public:
int compareVersion(string version1, string version2) {
if(version1.empty() && version2.empty()) return 0;
if(version1.empty()) version1 ="0";
if(version2.empty()) version2 ="0";
int index1 = version1.find(".");
int index2 = version2.find(".");
int v11 = index1 == -1 ? stoi(version1) : stoi(version1.substr(0, index1));
int v21 = index2 == -1 ? stoi(version2) : stoi(version2.substr(0, index2));
if(v11 > v21) return 1;
else if(v11 < v21) return -1;
else
{
string s1, s2;
if(index1 != -1) s1 = version1.substr(index1 1);
if(index2 != -1) s2 = version2.substr(index2 1);
return compareVersion(s1, s2);
}
}
};
