Compare Version Numbers

https://leetcode.com/problems/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

解题思路：

这种题目，面试的时候应该先问，version可能有多个'.'吗？ 还要再问，version里可能有某一部分为0吗？或者结尾连续都是0，比如13.1.0.0.0？否则只能埋着头写程序。

事实证明，这两种情况是都会有的，加入到考虑里也不难。我倒是怀疑这题考得是如何实现split()方法还是怎么的。

提前判断好两个split数组的大小，也可以做。

 

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] version1SPlit = version1.split("\\.");
        String[] version2SPlit = version2.split("\\.");
        int i = 0;
        while(i < version1SPlit.length && i < version2SPlit.length){
            if(Integer.parseInt(version1SPlit[i]) > Integer.parseInt(version2SPlit[i])){
                return 1;
            }
            if(Integer.parseInt(version1SPlit[i]) < Integer.parseInt(version2SPlit[i])){
                return -1;
            }
            i++;
        }
        if(i == version1SPlit.length && i == version2SPlit.length){
            return 0;
        }else if(i == version1SPlit.length){
            for(; i < version2SPlit.length; i++){
                if(Integer.parseInt(version2SPlit[i]) > 0){
                    return -1;
                }
            }
            return 0;
        }else{
            for(; i < version1SPlit.length; i++){
                if(Integer.parseInt(version1SPlit[i]) > 0){
                    return 1;
                }
            }
            return 0;
        }
    }
}