[Leetcode165] Compare Version Numbers

版本号比较算法

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本文介绍了一种用于比较两个版本号大小的算法，并提供了一个具体的Java实现案例。该算法能够正确处理版本号中数字序列的不同长度及尾部多余的零。

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

solution

细节实现题

if (no dot) directy compare,

else compare following digit

Need to take care of of the trailing 0 or non0s

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] versions1 = version1.split("\\.");
        String[] versions2 = version2.split("\\.");
        if(versions1.length==1 && versions2.length==1) return compare(version1, version2);
        int index = 0;
        while(index<versions1.length && index<versions2.length){
            int v1 = compare(versions1[index], versions2[index]);
            if(v1!=0) return v1;
            else {
                index++;
                continue;
            }
        }
        if(versions1.length > versions2.length){
            while(index<versions1.length){
                if(Integer.valueOf(versions1[index])!=0) return 1;
                else {
                    index++;
                    continue;
                }
            }
        } else{
            while(index<versions2.length){
                if(Integer.valueOf(versions2[index])!=0) return -1;
                else {
                    index++;
                    continue;
                }
            }
        }
        return 0;
    }
    public int compare(String t1, String t2){
        if(Integer.valueOf(t1)>Integer.valueOf(t2)) return 1;
        else if(Integer.valueOf(t1)<Integer.valueOf(t2)) return -1;
        else return 0;
    }
}