Find Single Number III

Given an integer array nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.

Example 1:

Input: nums = {1,2,1,3,2,5}
Output: [3,5]

Approach

Java

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class SingleNumberIII {
    public static void main(String[] args) {
        int nums[] = { 1, 2, 1, 3, 2, 5 };
        int arr[] = singleNumber(nums);
        System.out.println(Arrays.toString(arr));

    }

    static int[] singleNumber(int[] nums) {
        List<Integer> list = new ArrayList<Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (list.indexOf(nums[i]) != -1) {
                list.remove(list.indexOf(nums[i]));
            } else
                list.add(nums[i]);
        }
        return list.stream().mapToInt(Integer::intValue).toArray();
    }
}

C++

#include <bits/stdc++.h>
using namespace std;


vector<int> singleNumber(vector<int>& nums)
{
        set<int> st;
        for(int i=0;i<nums.size();i++)
        {
            if(st.find(nums[i])!=st.end())
            {
                st.erase(st.find(nums[i]));
            }
            else
                st.insert(nums[i]);
        }
        vector<int> res;
        for(auto it=st.begin();it!=st.end();it++)
               res.push_back(*it);
        return res;
}

int main()
{
    vector<int> nums ={1,2,1,3,2,5};
    vector<int> arr=singleNumber(nums);
    for(int i=0;i<arr.size();i++)
      cout<<arr[i]<<" ";
    return 0;
}