leetcode - Compare Version Numbers-优快云博客

本文详细阐述了如何通过编程实现版本号之间的比较，包括解析版本号字符串、处理点分隔符并进行数值比较，最终返回版本号间的相对大小。

leetcode - Compare Version Numbers

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

 1 class Solution {
 2 public:
 3     int compareVersion(string version1, string version2) {
 4         string::iterator it1;
 5         int ver1[1000], ver2[1000];
 6         memset(ver1,0,sizeof(ver1));
 7         memset(ver2,0,sizeof(ver2));
 8         it1=version1.begin();
 9         int i = 0;
10         
11         while(it1!=version1.end()){
12             ver1[i] = ver1[i]*10 + (*it1)-'0';
13             it1++;
14             if(*it1=='.') {i++;it1++;}
15         }
16 
17         
18         it1 = version2.begin();
19         int j = 0;
20         while(it1!=version2.end()){
21             ver2[j] = ver2[j]*10 + (*it1)-'0';
22             it1++;
23             if(*it1=='.') {j++;it1++;}
24         }
25 
26         for(int k=0; k<=max(i,j); k++ ){
27             if(ver1[k]>ver2[k]) return 1;
28             if(ver1[k]<ver2[k]) return -1;
29         }
30         return 0;
31         
32     }
33 };