Getting Factors of a Number

Question

I\'m trying to refactor this algorithm to make it faster. What would be the first refactoring here for speed?

public int GetHowManyFactors(int numberToCheck         


        

Answer 1

1. You can limit the upper limit of your FOR loop to numberToCheck / 2

2. Start your loop counter at 2 (if your number is even) or 3 (for odd values). This should allow you to check every other number dropping your loop count by another 50%.

   public int GetHowManyFactors(int numberToCheck)
   {
     // we know 1 is a factor and the numberToCheck
     int factorCount = 2; 
   
     int i = 2 + ( numberToCheck % 2 ); //start at 2 (or 3 if numberToCheck is odd)
   
     for( ; i < numberToCheck / 2; i+=2) 
     {
        if (numberToCheck % i == 0)
           factorCount++;
     }
     return factorCount;
   }