Compare Version Numbers

本文详细解释了如何比较两个版本号,并提供了相应的算法实现。包括版本号的解析、比较规则及示例应用。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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.

You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note:

  1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
  2. Version strings do not start or end with dots, and they will not be two consecutive dots.

思路:split注意是“\\.”,剩下的就跟merge two sorted list一样,同样的指针同时走;

class Solution {
    public int compareVersion(String version1, String version2) {
        if(version1.equals(version2)) {
            return 0;
        }
        String[] splits1 = version1.split("\\.");
        String[] splits2 = version2.split("\\.");
        int i = 0;
        while(i < splits1.length || i < splits2.length) {
            if(i < splits1.length && i < splits2.length) {
                int n1 = Integer.parseInt(splits1[i]);
                int n2 = Integer.parseInt(splits2[i]);
                if(n1 > n2) {
                    return 1;
                } else if(n1 < n2) {
                    return -1;
                }
            } else if(i < splits1.length) {
                if(Integer.parseInt(splits1[i]) != 0) {
                    return 1;
                }
            } else if(i < splits2.length) {
                if(Integer.parseInt(splits2[i]) != 0) {
                    return -1;
                }
            }
            i++;
        }
        return 0;
    }
}

 

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值