[LeetCode] Compare Version Numbers 字符串操作

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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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  这题是太多细节的操作,字符串string 的操作其实很简单,主要判断0的情况,需要考虑1.1.0 与1.1 是相等的。代码写的有点冗余,如果再第一个while 中修改可以改短。
 
#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
    int compareVersion(string version1, string version2) {
        if (version1.length()<1||version2.length()<1)   return 0;
        int beg1 = 0,beg2 = 0;
        int end1 = 0;
        int end2 = 0;

        while(end1!=string::npos&&end2!=string::npos){
            end1 = version1.find_first_of('.',beg1);
            end2 = version2.find_first_of('.',beg2);
            int int1 = hFun(version1.substr(beg1,end1-beg1));
            int int2 = hFun(version2.substr(beg2,end2-beg2));
            if(int1 > int2) return 1;
            if(int1 < int2) return -1;
            beg1 = end1 +1;
            beg2 = end2 +1;
        }
        if(end1==string::npos){
            while(end2!=string::npos){
                end2 = version2.find_first_of('.',beg2);
                int int2 = hFun(version2.substr(beg2,end2-beg2));
                if(int2 >0) return -1;
                beg2 = end2+1;
            }
        }
        if(end2==string::npos){
            while(end1!=string::npos){
                end1 = version1.find_first_of('.',beg1);
                int int1 = hFun(version1.substr(beg1,end1-beg1));
                if(int1 >0) return 1;
                beg1 = end1+1;
            }
        }
        return 0;
    }
    int hFun(string s)
    {
        int n = 0;
        for(int i=0;i<s.length();i++){
            n*=10;
            n+=s[i]-'0';
        }
        return n;
    }
};

int main()
{
    string version1 = "1.1";
    string version2 = "1";
    Solution sol;
    cout<<sol.compareVersion(version1,version2)<<endl;
    return 0;
}
 
 
 
  
原文地址:https://www.cnblogs.com/Azhu/p/4323837.html