康瑞部落

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

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

解题思路：

这道题思路倒是挺简单的，按每个‘.’进行分割，逐个比较大小即可。但是要考虑不是同样多个‘.’和1.0与1.0.0等情况。

代码如下：

class Solution {
public:
    int compareVersion(string version1, string version2) {
        string leftVersion1 = version1;
        string leftVersion2 = version2;
        while(leftVersion1!=""&&leftVersion2!=""){
            int subver1, subver2;
            
            int pos1 = leftVersion1.find('.');
            if(pos1!=string::npos){
                subver1 = strToInt(leftVersion1.substr(0, pos1));
                leftVersion1 = leftVersion1.substr(pos1 + 1);
            }else{
                subver1 = strToInt(leftVersion1);
                leftVersion1 = "";
            }
            
            int pos2 = leftVersion2.find('.');
            if(pos2!=string::npos){
                subver2 = strToInt(leftVersion2.substr(0, pos2));
                leftVersion2 = leftVersion2.substr(pos2 + 1);
            }else{
                subver2 = strToInt(leftVersion2);
                leftVersion2 = "";
            }
            
            if(subver1>subver2){
                return 1;
            }
            if(subver1<subver2){
                return -1;
            }
        }
        if(leftVersion1!=""&&!checkAllZero(leftVersion1)){
            return 1;
        }
        if(leftVersion2!=""&&!checkAllZero(leftVersion2)){
            return -1;
        }
        return 0;
    }
private:
    int strToInt(string s){
        int base = 1;
        int result = 0;
        int len = s.length();
        for(int i=len-1;i>=0; i--){
            result += base*(s[i]-'0');
            base *= 10;
        }
        return result;
    }
    bool checkAllZero(string s){
        int len = s.length();
        for(int i = 0; i<len; i++){
            if(s[i]!='.' && s[i]!='0'){
                return false;
            }
        }
        return true;
    }
};

二次刷题（2015-08-05）

这道题挺无聊的~

class Solution {
public:
    int compareVersion(string version1, string version2) {
        while(version1!=""&&version2!=""){
            int pos1 = version1.find('.');
            int pos2 = version2.find('.');
            int subversion1, subversion2;
            if(pos1<0){
                subversion1 = atoi(version1.c_str());
                version1 = "";
            }else{
                string s1 = version1.substr(0, pos1);
                subversion1 = atoi(s1.c_str());
                version1 = version1.substr(pos1 + 1);
            }
            if(pos2<0){
                subversion2 = atoi(version2.c_str());
                version2 = "";
            }else{
                string s2 = version2.substr(0, pos2);
                subversion2 = atoi(s2.c_str());
                version2 = version2.substr(pos2 + 1);
            }
            if(subversion1<subversion2){
                return -1;
            }else if(subversion1>subversion2){
                return 1;
            }
        }
        if(version1!=""&&!checkAllZero(version1)){
            return 1;
        }
        if(version2!=""&&!checkAllZero(version2)){
            return -1;
        }
        return 0;
    }
    
    bool checkAllZero(string s){
        int len = s.length();
        for(int i = 0; i<len; i++){
            if(s[i]!='.' && s[i]!='0'){
                return false;
            }
        }
        return true;
    }
};