LeetCode -- Compare Version Numbers

题目描述：

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

思路：
1.把用'.'隔开的版本号存入数组中，存为arr1和arr2
2.同时遍历arr1和arr2，比较arr1[i]和arr2[i]的大小，如果不等：如果arr1[i]>arr2[i]返回1，否则返回-1。 在比较时需要注意版本0、以及其中一个版本为空的情况
3.遍历到最后没有发现不等，返回0（相等）

实现代码：

public class Solution {
public int CompareVersion(string version1, string version2) {
var v1 = version1;
var v2 = version2;

var v1Arr = v1.Split('.');
var v2Arr = v2.Split('.');

var i = 0;
var j = 0;
while(i < v1Arr.Length || j < v2Arr.Length){
if(i == v1Arr.Length){
return int.Parse(v2Arr[i]) == 0 ? 0 : -1;
}
if(i == v2Arr.Length){
return int.Parse(v1Arr[i]) == 0 ? 0 : 1;
}

if(int.Parse(v1Arr[i]) == int.Parse(v2Arr[j])){
i++;
j++;
continue;
}

return int.Parse(v1Arr[i]) > int.Parse(v2Arr[j]) ? 1 : -1;
}

return 0;
}
}