1、题目

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

2、wepon的解法

2.1 解析

2.2 代码

C++ class Solution { public: int compareVersion(string version1, string version2) { vector result1=getInt(version1); vector result2=getInt(version2); int len1=result1.size(); int len2=result2.size(); if(len2<len1) return -1*compareVersion(version2, version1); int i=0; while(i<len1 && result1[i]==result2[i]) i++;

        if(i==len1){    //str1和str2前len1位都相等，则看看str2后面的len2-len1位是否都为0即可判断它们的大小
            int j=len2-1;
            while(j >= len1){
                if(result2[j--]!=0) return  -1;
            }
            return 0;
        }else{       //str1和str2前len1位不都相等，直接判断第i位
            if(result1[i]<result2[i]) return -1;
            else return 1;
        }
    }
private:
//将version字符串按'.'拆成多个，转化为整型放入容器
    vector<int> getInt(string version){
        vector<int> result;
        int len=version.size();
        int pre=0;
        for(int i=0;i<len;i++){
            if(version[i]=='.'){
                string str(version.begin()+pre,version.begin()+i);  //注意这种初始化形式，左闭右开，即str不包括version[version.begin()+i]
                result.push_back(stoi(str));
                pre=i+1;
            }
        }
        string str(version.begin()+pre,version.end());
        result.push_back(stoi(str));
        return result;
    }
};

3、有更好的解法？快来占位吧 ^_^