*(leetcode_String) Compare Version Numbers

Compare Version Numbers

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Question  Solution 

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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比较繁琐，要考虑清楚各种情况  divede函数可以用strtok函数~ 

class Solution {
    vector<int> divide(string version){
        vector<int> ret;
        int len = version.length();
        int preIndex = 0;
        for(int i=0; i<len; i++){
            if(version[i]=='.'){
                string str = version.substr(preIndex, i-preIndex);
                ret.push_back(atoi(str.c_str()));
                preIndex=i+1;
            }
        }
        string str = version.substr(preIndex, len+1-preIndex);
        ret.push_back( atoi(str.c_str()) );
        return ret;
    }
public:
    int compareVersion(string version1, string version2) {
        vector<int> v1 = divide(version1);
        vector<int> v2 = divide(version2);
        int i=0,j=0;
        while(i<v1.size()&&j<v2.size()){
            if(v1[i]>v2[j])
                return 1;
            if(v1[i]<v2[j])
                return -1;
            ++i;
            ++j;
        }
        if(i==v1.size()&&j<v2.size()){
            while(j<v2.size()) //后面的数字也要>0， 忽略结尾的0
                if(v2[j++]>0)
                    return -1;
        }
        if(i<v1.size()&&j==v2.size()){
            while(i<v1.size())
                if(v1[i++]>0)
                    return 1;
        }
        return 0;
    }
};