Leetcode题解 165. Compare Version Numbers

最新推荐文章于 2022-07-18 13:37:52 发布

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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实现。通过解析版本号字符串中的数字序列，并逐位对比来确定版本间的先后顺序。该算法适用于不同层级修订版本的比较。

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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

注意：split(“\.”) 参数’.’要转义，因为split的参数是正则表达式，而’.’代表匹配任意字符

public class Solution {
    public int compareVersion(String version1, String version2) {
        int count1=0;
        int count2=0;
        for(int i=0;i<version1.length();i++){
            if(version1.charAt(i)=='.'){
                count1++;
            }
        }
        for(int i=0;i<version2.length();i++){
            if(version2.charAt(i)=='.'){
                count2++;
            }
        }
        if(count1==0&&count2==0){
            if(Integer.parseInt(version1)>Integer.parseInt(version2)){
                return 1;
            }else if(Integer.parseInt(version1)<Integer.parseInt(version2)){
                return -1;
            }
            return 0;
        }
        if(count1>count2){
            int diff=count1-count2;
            while(diff>0){
                version2+=".0";
                diff--;
            }
        }else if(count2>count1){
            int diff=count2-count1;
            while(diff>0){
                version1+=".0";
                diff--;
            }
        }
        String[] res1=version1.split("\\.");
        String[] res2=version2.split("\\.");
        for(int i=0;i<res1.length;i++){
            if(Integer.parseInt(res1[i])>Integer.parseInt(res2[i])){
                return 1;
            }else if(Integer.parseInt(res1[i])<Integer.parseInt(res2[i])){
                return -1;  
            }
        }
        return 0;
    }
}