260. Single Number III

Problem:

My Submissions Total Accepted: 56553 Total Submissions: 115017 Difficulty: Medium Contributors: Admin Given an array of numbers 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.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note: The order of the result is not important. So in the above example, [5, 3] is also correct. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Solutions:

public class Solution {
    public int[] singleNumber(int[] nums) {
        HashSet<Integer> app = new HashSet<Integer>();
        for (int i = 0; i < nums.length; i ++) {
            if (app.contains(nums[i])) {
                app.remove(nums[i]);
            }
            else {
                app.add(nums[i]);
            }
        }
        int[] result = new int[2];
        Iterator<Integer> it = app.iterator();
        result[0] = it.next();
        result[1] = it.next();
        return result;
    }
}

public class Solution {
    public int[] singleNumber(int[] nums) {
        int ab = 0;
        for (int i = 0; i < nums.length; i ++) {
            ab = ab ^ nums[i];
        }
        int oneIndex = 0;
        for (int i = 0; i < 32; i ++) {
            if (((ab >>> i) & 1) == 1) {
                oneIndex = i;
                break;
            }
        }
        int[] result =  new int[2];
        result[0] = 0;
        for (int i = 0; i < nums.length; i ++) {
            if (((nums[i] >>> oneIndex) & 1) == 1) {
                result[0] = result[0] ^ nums[i];
            }
        }
        result[1] = ab ^ result[0];
        return result;
    }
}