[LeetCode] 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


该题的主要意图在于考察思考问题的全面性,该题体现了C/C++的性能优势,java代码贴出.

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] sv1 = version1.split("\.");
        String[] sv2 = version2.split("\.");
        int idx1 = 0;
        int idx2 = 0;
        
        while(idx1 < sv1.length || idx2 < sv2.length){
            int v1 = 0;
            if(idx1 < sv1.length){
                v1 = Integer.parseInt(sv1[idx1]);
                idx1++;
            }
            int v2 = 0;
            if(idx2 < sv2.length){
                v2 = Integer.parseInt(sv2[idx2]);
                idx2++;
            }
            if(v1 < v2)
                return -1;
            if(v1 > v2)
                return 1;
        }
        return 0;
    }
}
原文地址:https://www.cnblogs.com/HackingProgramer/p/4354757.html