Ggtcf

I need to write a method of the sum of the numbers that are between the two given integers.

For example, the call sumNums(1,3) should return 6 because 1+2+3 = 6.

The call sumNums(3,1) should also return 6.

So far I have

public static int sumNum(int num1, int num2)

 int min = Math.min(num1, num2);
 int max = Math.max(num1, num2);
 int sum = 0;
 for(int i = min; i < max; i++)
 
 sum += i;
 
 sum =+ max;
 return sum;

any other methods would help, but if there's a way initialize the sum and return it it would help.

public static int sumNum(int num1, int num2) 
 int min = Math.min(num1, num2);
 int max = Math.max(num1, num2);
 int sum = 0;
 for (int i = min; i <= max; i++) 
 sum += i;
 
 return sum;
 

Honestly using java-8 IntStream

IntStream.rangeClosed(1, 3).sum(); //6

using java-7 and below

public static int sumNum(int num1, int num2)
 
 int min = Math.min(num1, num2);
 int max = Math.max(num1, num2);
 int sum = 0;
 for(int i = min; i <= max; i++)
 
 sum += i;
 

 return sum;

It's (simple) math:

public static int sumNum(int num1, int num2) 
 int min = Math.min(num1, num2);
 int max = Math.max(num1, num2);
 return max * (max + 1) / 2 - min * (min - 1) / 2

You could do something likes this:

return (num1 + num2) / 2.0 * (Math.abs(num1 - num2) + 1);

Your current approach will encounter performance problems the larger the gap between the two integers. There are optimisations you can make.

For example, if you have an odd number of integers:

2 + 3 + 4 + 5 + 6 = 4 + 4 + 4 + 4 + 4

1 + 2 + 3 = 2 + 2 + 2

5 + 6 + 7 = 6 + 6 + 6

See the pattern?

2 + 3 + 4 + 5 + 6 = 4 + 4 + 4 + 4 + 4
 = 5 * 4
 = (max - min + 1) * ((max + min) / 2)

For an even number of integers:

2 + 3 + 4 + 5 = 3.5 + 3.5 + 3.5 + 3.5
 = 4 * 3.5

5 + 6 + 7 + 8 + 9 + 10 = 7.5 + 7.5 + 7.5 + 7.5 + 7.5 + 7.5
 = 6 * 7.5

Looking familiar! Let's see if our previous formula works:

2 + 3 + 4 + 5 = (max - min + 1) * ((max + min) / 2)
 = (5 - 2 + 1) * ((5 + 2) / 2)
 = 4 * 3.5

The formula can be simplified:

(max - min + 1) * ((max + min) / 2) = (max - min + 1) * (max + min) / 2

Therefore:

public static int sumNum(int num1, int num2) 
 int min = Math.min(num1, num2);
 int max = Math.max(num1, num2);
 return (max - min + 1) * (max + min) / 2;

Or using math to get it in constant time:

sum = max * (max + 1) / 2 - (min * (min - 1)) /2;