How to find all odd occurring elements in an array in JAVA -

i trying find elements occur odd number of times in array. worked out little bit, code returning correct answer if there 1 number occurs odd number of times. if there 2 or more odd occurring numbers not able process it. understand if bit-wise xor of elements 1 odd occurring element. how can improve multiple numbers?

below code:

public class oddoccur {     public int oddoccurence(int[] arr, int size) {         int res = 0;         int[] fin = new int[size];          (int = 0; < size; i++) {             res = res ^ arr[i];          }         return res;     }      public static void main(string args[]) {         int[] arr = { 2, 5, 5, 2, 2, 3, 3, 3 };          int n = arr.length;         oddoccur obj = new oddoccur();          system.out.println("odd occuring element is:"                 + obj.oddoccurence(arr, n));     } } 

need solve issue!

public int oddoccurence(int[] arr, int size); 

first, there can multiple numbers occur odd number of times. there's no way write function , make work if returns single int. you'll need function can return multiple numbers.

public int[] oddoccurrences(int[] array); 

second, don't try clever. using xor "clever". you're not going find simple solution using xor. it'll convoluted, complicated mess. keep simple:

1. count number of times each number appears.
2. then find odd counts.

example pseudo-code:

// map numbers counts. assume counts start @ 0. map<integer, integer> counts;  (number in array) {     counts[number] += 1 }  // list of numbers occur odd number of items. list<integer> oddnumbers;  (number, count in counts) {     if (count % 2 != 0) {         oddnumbers.add(number);     } }