checking an array of ints to see if the sum of any two elements equals another

Question

Before anyone says it...yes...this is a homework assignment. Basically, I am going to be given an array of ints and I am suppose to write a program that accepts the array and then checks it to see if any two elements sum up to equal a third one.

So if the array is [1 2 3 4] then it's a "yes" because 1+2=3 and 1+3=4 but if it's [1 1 1 1] then it's a "no".

I've already written all other methods, I just need the logic to do what I explained above, preferably using recursion (because that's the assignment).

I honestly don't even know where to start with recursion, never used it before.

Answer 1

You could use this, to find the sum of every two numbers in the array, and then compare them with the array's contents.

 for(int i = 0; i < array.length; i++) {
        for(int j = 0; j < array.length; j++) {
            int sum = array[i] + array[j];
            for(int k = 0; k < array.length; k++) {
                if(array[k] == sum)
                    return true;
                else
                    return false;
            }
        }
    }

This may not be the most efficient, but it works, and that is what you were looking for.

Answer 2

Here I have used Integer[]

Integer[] arr = new Integer[] { 1, 2, 3, 4 };
for (int i = 0; i < arr.length - 1; i++) {
    if(arr[i] == 0) {
         continue;
    }
    for (int k = i + 1; k < arr.length; k++) {
        if (arr[k] != 0
                && Arrays.asList(arr).contains(arr[i] + arr[k])) {
            return true;
        }
    }
}
return false;

Arrays.asList(arr) converts arr to a list and contains(arr[i] + arr[k]) will check whether the sum of the 2 elements are present in arr

Answer 3

You could always use a for loop like this:

for(int i = 0; i+2 < array.length; i++) //i + 2 because the last two element pair won't be used
        if(array[i+2] == array[i] + array[i+1]){return true;} //Or whatever else you want to return
        else{return false;}
}