Compare Version Numbers Leetcode

最新推荐文章于 2019-09-06 20:06:31 发布

转载最新推荐文章于 2019-09-06 20:06:31 发布 · 63 阅读

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原文链接：http://www.cnblogs.com/aprilyang/p/6343219.html

本文提供了一种有效的版本号比较算法实现。通过将版本号字符串分解并进行逐级对比，能够准确判断两个版本号之间的大小关系。该算法适用于软件版本管理和更新检查等场景。

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

反正看完人家的就觉得自己的改来改去还是一坨翔。。。= =

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");
        int index = 0;
        while (index < v1.length && index < v2.length) {
            int a = Integer.valueOf(v1[index]);
            int b = Integer.valueOf(v2[index]);
            if (a - b < 0) {
                return -1;
            } else if (a - b > 0) {
                return 1;
            }
            index++;
        }
        if (index < v1.length) {
            int a = Integer.valueOf(v1[index]);
            while (a == 0) {
                index++;
                if (index < v1.length) {
                    a = Integer.valueOf(v1[index]);
                } else {
                    return 0;
                }
            }
            if (index < v1.length) {
                return 1;
            }
        }
        if (index < v2.length) {
            int b = Integer.valueOf(v2[index]);
            while (b == 0) {
                index++;
                if (index < v2.length) {
                    b = Integer.valueOf(v2[index]);
                } else {
                    return 0;
                }
            }
            if (index < v2.length) {
                return -1;
            }
        }
        return 0;
    }
}

看完top solution的答案是这样的

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");

        int length = Math.max(v1.length, v2.length);
        for (int i = 0; i < length; i++) {
            Integer a = i < v1.length ? Integer.valueOf(v1[i]) : 0;
            Integer b = i < v2.length ? Integer.valueOf(v2[i]) : 0;
            int compare = a.compareTo(b);
            if (compare != 0) {
                return compare;
            }
        }
        return 0;
    }
}

?差距。。。这个方法主要在于，不够长的用0来补长，这样就可以省去很多判断直接减。另外用Integer作为类，然后调用compareTo的方法很好。

转载于:https://www.cnblogs.com/aprilyang/p/6343219.html