## Compare Version Numbers

07/15/2016 String

## 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={0}, v1n = 0;
int v2={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: $O(1)$