带返回的递归方法

Question

I have to write a recursive method that will count the sum of all even numbers that can not be divided with 7 in an interval form 1 till a parameter.

I cant use any loop. This is how far I came, but it seems that my statement is not correct.

Any suggestions?

public static int specialSum (int x){
    return (x % 2 == 0) && (x % 7 != 0) ? specialSum(x - 1) + 1 : 0;
}

public static void main(String[] args) {
    System.out.println(specialSum(16));
}

Answer 1

You need to recurse to the next number whether or not you count the current one.

boolean divisibleBy2Not14 = ((x % 2 == 0) && (x % 7 != 0);
return (divisibleBy2Not14 ? x : 0) + (x > 0 ? specialSum(x - 1) : 0);

Answer 2

If you need to find the sum of such ( x % 2 == 0 && x % 7 != 0 ) positive ( x > 0 ) numbers:

public static int specialSum (int x) {
    return x > 0 ?
            specialSum(x - 1) + (x % 2 == 0 && x % 7 != 0 ? x : 0) :
            0;
}

Answer 3

The logic of your recursive has two problems:

1. It's trying to return the count and not the sum of number that are valid
2. And the recursion does not reach all the branches, because it's terminated as soon as it reaches a non-valid case.

If you want the sum you should return specialSum(x-1)+x not specialSum(x-1)+1 . This is an example that would work:

public static int specialSum (int x){
    if(x == 0) return 0; // ← degenerate case
    return ((x % 2 == 0) && (x % 7 != 0) ? specialSum(x - 1) + x : specialSum(x - 1));
}

You could add a little clever simplification by replacing specialSum(x - 1) + x by specialSum(x - 2) + x , because you know that x - 1 is going to be odd if x is even.

Answer 4

you have to change the specialsum method like this:

public static int specialSum (int x){
    if(x == 1)  return 0;
    return ((x % 2 == 0) && (x % 7 != 0) ? specialSum(x - 1) + x   : specialSum(x - 1));
}

Answer 5

A good way to approach it is to first write it out the long way (no ternary operators). Then you can see if it can be shortened:

   public static int summer( int n ) {
      if ( n < 2 ) {
         return 0;
      } else if ( (n % 2) == 0 ) {
         if ( (n % 7) == 0 ) {
            return summer( n - 2 ); // n-1 is odd, so skip it
         }
         return n + summer( n - 2 ); // n-1 is odd, so skip it
      } else {
         return summer( n - 1 );
      }
   }