Compare Version Numbers

07/15/2016 String

Question
Solution

Question

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

Solution

Result: Accepted Time: 0 ms

Here should be some explanations.

int compareVersion(char* version1, char* version2) {
    int v1[500]={0}, v1n = 0;
    int v2[500]={0}, v2n = 0;
    int n1 = 0, n2 = 0;
    while(*version1)
    {
        if(*version1 == '.')
        {
            v1[v1n++] = n1;
            n1 = 0;
        }
        else
        {
            n1 = n1*10 + *version1 - '0';
        }
        ++version1;
    }
    v1[v1n++] = n1;
    while(*version2)
    {
        if(*version2 == '.')
        {
            v2[v2n++] = n2;
            n2 = 0;
        }
        else
        {
            n2 = n2*10 + *version2 - '0';
        }
        ++version2;
    }
    v2[v2n++] = n2;
    int i1=0,i2 =0;
    while(i1 < v1n || i2 < v2n)
    {
        if(v1[i1] > v2[i2])
            return 1;
        else if(v1[i1] < v2[i2])
            return -1;
        ++i1,++i2;
    }
    return 0;
}

Complexity Analytics

Time Complexity: $O(n)$
Space Complexity: $O(1)$