LeetCode -- Compare Version Numbers

LeetCode -- Compare Version Numbers

題目描述：


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'.'隔開的版本號存入數組中，存為arr1和arr2
2.同時遍歷arr1和arr2，比較arr1[i]和arr2[i]的大小，如果不等：如果arr1[i]>arr2[i]返回1，否則返回-1。 在比較時需要注意版本0、以及其中一個版本為空的情況
3.遍歷到最後沒有發現不等，返回0（相等）




實現代碼：

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;
    }
}