recursively sum the integers in an array

Question

I have a program that I\'m trying to make for class that returns the sum of all the integers in an array using recursion. Here is my program thus far:

public         


        

Answer 1

private static int sum(int[] arr) {
    // TODO Auto-generated method stub
    int n = arr.length;

    if(n==1)
    {
        return arr[n-1];
    }

    int ans = arr[0]+sum(Arrays.copyOf(arr, n-1));

    return ans;
}

Answer 2

a is an int array. Thus a[n-1] is an int. You are passing an int to sumOfArray which expects an array and not an int.

Answer 3

Simplified version:

//acc -> result accumlator, len - current length of array

public static int sum(int[] arr, int len, int acc) {
    return len == 0 ? acc :  sum(arr, len-1,  arr[len-1]+ acc); 
}   
public static void main(String[] args)  {
    int[] arr= { 5, 1, 6, 2};
    System.out.println(sum(arr, arr.length, 0));
}

Answer 4

How about this recursive solution? You make a smaller sub-array which contains elements from the second to the end. This recursion continues until the array size becomes 1.

import java.util.Arrays;

public class Sum {
    public static void main(String[] args){
        int[] arr = {1,2,3,4,5};
        System.out.println(sum(arr)); // 15
    }

    public static int sum(int[] array){
        if(array.length == 1){
            return array[0];
        }

        int[] subArr = Arrays.copyOfRange(array, 1, array.length);
        return array[0] + sum(subArr);
    }
}

Answer 5

The solution is simpler than it looks, try this (assuming an array with non-zero length):

public int sumOfArray(int[] a, int n) {
    if (n == 0)
        return a[n];
    else
        return a[n] + sumOfArray(a, n-1);
}

Call it like this:

int[] a = { 1, 2, 3, 4, 5 };
int sum = sumOfArray(a, a.length-1);

Answer 6

a[n-1] 

is getting the int at n-1, not the array from 0 to n-1.

try using

Arrays.copyOf(a, a.length-1);

instead