在数组中找到总和等于给定数字的所有数字对？

Question

Write a program that passes in an array and will then output:

1) all pairs that sum up to 10. 2) all unique pairs that sum to 10, no duplicates. 3) all unique pairs that sum to 10, no duplicates or reverse pairs.

I have the bellow so far but it only prints the following:
(1, 9) (1, 9) (4, 6) (4, 6) (5, 5) (5, 5)

public class SumOfPairs { 

public void pairedElements(int arr[], int sum) 
{ 
    int low = 0; 
    int high = arr.length -1; 

    while (low < high) { 
        if (arr[low] + arr[high] == sum) { 
            System.out.println(" ("+ arr[low] + ", " + arr[high] + ")");
        } 
        if (arr[low] + arr[high] > sum) { 
            high--; 
        } 
        else { 
            low++; 
        } 
    } 
} 

public static void main(String[] args) 
{ 
    int arr[] = {1, 1, 2, 4, 4, 5, 5, 5, 6, 7, 9}; 
    Arrays.sort(arr); 

    SumOfPairs sp = new SumOfPairs(); 
    sp.pairedElements(arr, 10); 
} 

Answer 1

You can use nested for loops. In outer loop you can iterate from 0 up to the last but one. In inner loop you can iterate from outer index + 1 up to the last element. Something like this:

for(int i = 0; i < arr.length -1; i++){
   for(int j = i + 1; j < arr.length; j++){
      if(arr[i] + arr[j] == 10){
        //do something
      }
   }
}

Answer 2

As stated above use a nested for loop to find your pairs. To find duplicates or Unique pairs i would store them in a list and check the list for instances of a value.

public void pairedElement(int [] arr, sum){
List<String> pairs = new ArrayList<>();

for(int x : arr){
   for(int y : arr){
     if(x + y == sum)
      pairs.add("(" + x + "," + y + ")");
   }
}
 //to print all of the values:
pairs.foreach(x -> System.out.println(x));

To print the unique/duplicate pairs only id recommend checking out this Specifically the second answer will give you an idea of how to get 2 collections, one containing the unique items from the entire list (pairs) and the other containing values that are duplicates in the original list.