[LeetCode]Compare Version Numbers

最新推荐文章于 2016-12-18 10:34:10 发布

原创最新推荐文章于 2016-12-18 10:34:10 发布 · 420 阅读

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本文详细介绍了如何比较两个版本号之间的大小关系，并通过代码实现。重点在于理解版本号的组成与转换策略，以及如何处理小数部分的比较。

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

LeetCode Source

分析：写的比较乱，这个题目不难，注意好边界条件。把version转换成了大版本int型m，和小版本float型n。

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int m1=0;
        float n1=0;
        int m2=0;
        float n2=0;
        ToInt(version1,m1,n1);
        ToInt(version2,m2,n2);
        if(m1>m2)
            return 1;
        if(m1<m2)
            return -1;
        if(m1==m2){
            if(n1>n2)   return 1;
            if(n1<n2)   return -1;
            if(n1==n2)  return 0;
        }   
    }
    void ToInt(string s,int &m,float &n){
        int sz=s.size();
        int flag = 0;
        int i = 0;
        for(;i<sz;++i){
            if(s[i]=='.') break;
            else{
                m = m*10+s[i]-'0';
            }
        }
        for(i=i+1;i<sz;++i){
            if(s[i]=='.') {
                if(s[i+1]!='0')
                    flag ++;
                if(s[i+1]=='0'&&i+2>sz-1)
                    break;
                continue;
            }
            n = n*10+s[i]-'0';
        }
        while(flag>0){
            n=n/10;
            --flag;
        }
    }
};