Multiply 2 long numbers without using multiplication.

Question

Saravana Meiyappan · Accepted Answer

- (NSInteger)multiplyNum:(NSInteger)num1 with:(NSInteger)num2 {
  
  if (num2 == 0 )
     return 0;
  
  if (num2 > 0)
  return (num1 + [self multiplyNum:num1 with:num2-1]); 
  
  if (num2 < 0)
  return -[self multiplyNum:num1 with:-num2];  
  
  return 0;
}

Utilisateur anonyme · Answer

Bit wise operation is a possibility along with addition. I wonder if this question is more about thinking outside of the box than coming up with the most efficient alternative.

Alex Darwish · Answer

def multiply(num1,num2):
    results=0
    results_array=[]
    for i in range (1,num2+1):
        results += num1 
    print(results)

a=5
b=5
multiply(a,b)

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