165. Compare Version Numbers

最新推荐文章于 2024-06-04 07:48:41 发布

原创最新推荐文章于 2024-06-04 07:48:41 发布 · 196 阅读

0 ·

CC 4.0 BY-SA版权

leetcode代码笔记专栏收录该内容

82 篇文章

订阅专栏

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并逐段比较数字，该算法能准确判断出两个版本号之间的大小关系。

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

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int n = version1.size(), m = version2.size();
        int i = 0, j = 0;
        int num1 = 0, num2 = 0;
        while(i < n || j < m){
        while(i < n){
            if(version1[i] != '.'){
                num1 = num1*10 + (version1[i]-'0');
                
                ++i;
            }
            else
            break;
        }
        while(j < m){
            if(version2[j] != '.'){
                num2 = num2*10 + (version2[j]-'0');

                ++j;
            }
            else
            break;
        }
        if(num1 > num2) return 1;
        if(num1 < num2) return -1;
        num1 = 0, num2 = 0;
        ++i;
        ++j;

        }
        return 0;
    }
    
};

还有很多用C++语言特性来解答的，之后再看