【leetcode】Compare Version Numbers

Compare Version Numbers

Compare two version numbers version1 and version1.
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.

 

 

考虑以下几种情况即可：

Example1: version1=="11.22.33", version2=="11.22.22". 11 == 11; 22 == 22; 33 > 22; return 1.

Example2: version1=="11.22.33", version2=="11.22.33". 11 == 11; 22 == 22; 33 == 33; return 0.

Example3: version1=="11.22.33", version2=="11.22.33.00.00". 11 == 11; 22 == 22; 33 == 33; remaining version2 equals to 0; return 0.

Example4: version1=="11.22.33.00.01", version2=="11.22.33". 11 == 11; 22 == 22; 33 == 33; remaining version1 contains 01; return 1.

 

 

第一种方法，把字符串拆分到vector中，然后进行对比：

class Solution {
public:
 
    
    void split(std::string& s, std::string& delim,std::vector< std::string >& ret)
    {
     size_t last = 0;
        size_t index=s.find_first_of(delim,last);
     while (index!=std::string::npos)
     {
      ret.push_back(s.substr(last,index-last));
      last=index+1;
      index=s.find_first_of(delim,last);
     }
     if (index-last>0)
     {
      ret.push_back(s.substr(last,index-last));
     }
    }
   
    int compareVersion(string version1, string version2) {
        
        //拆分到vector中
        vector<string> v1,v2;
        string delim=".";
       
        split(version1,delim,v1);
        split(version2,delim,v2);
       
        int n=v1.size()>v2.size()?v1.size():v2.size();
       
       
        for(int i=0;i<n;i++)
        {
 
            //当v1完了，只需要看v2末尾是否都为0即可
            if(i>=v1.size())
            {
                if(atoi(v2[i].c_str())!=0)
                {
                    return -1;
                }
            }
           
            //当v2完了，只需要看v1末尾是否都为0即可
            if(i>=v2.size())
            {
                if(atoi(v1[i].c_str())!=0)
                {
                    return 1;
                }
            }
           
            //v1,v2没完，则比较大小即可
            if(i<v1.size()&&i<v2.size())
            {
                if(atoi(v1[i].c_str())>atoi(v2[i].c_str()))
                {
                    return 1;
                }
                if(atoi(v1[i].c_str())<atoi(v2[i].c_str()))
                {
                    return -1;
                }
            }
        }
        return 0;
    }
};

 

 

第二种解法，采用库函数

C++ 11中，stoi可以从字符串string中取得数字，并返回不是数字下一个位置。

string中的erase函数：string& erase (size_t pos = 0, size_t len = npos)

 

class Solution {
public:
 
    int compareVersion(string version1, string version2) {
 
 
        while(!(version1.empty()&&version2.empty()))
        {
            size_t pos1=0;
            size_t pos2=0;
            int n1,n2;
 
            n1=version1.empty()?0:stoi(version1,&pos1);
            n2=version2.empty()?0:stoi(version2,&pos2);
 
           
            if(n1>n2)
                return 1;
            else if(n1<n2)
                return -1;
           
           
            if(!version1.empty())
                version1.erase(0,pos1+1);
            if(!version2.empty())
                version2.erase(0,pos2+1);
        }
       
        return 0;
       
    }
};