Leetcode – Single Number (Java)

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The problem:

Given an array of integers, every element appears twice except for one. Find that single one.

Java Solution 1

The key to solve this problem is bit manipulation. XOR will return 1 only on two different bits. So if two numbers are the same, XOR will return 0. Finally only one number left.

public int singleNumber(int[] A) {
	int x = 0;
	for (int a : A) {
		x = x ^ a;
	}
	return x;
}

Java Solution 2

public int singleNumber(int[] A) {
	HashSet<Integer> set = new HashSet<Integer>();
	for (int n : A) {
		if (!set.add(n))
			set.remove(n);
	}
	Iterator<Integer> it = set.iterator();
	return it.next();
}

The question now is do you know any other ways to do this?

12 thoughts on “Leetcode – Single Number (Java)”

  1. Java 8 version for XOR method.

    public static int singleNumber8(int[] elements)
    {
    return Arrays.stream(elements).reduce(0,(a, b) -> a ^ b);
    }


  2. public int singleNumber(int[] A){
    List list=Arrays.stream(A).boxed().collect(Collectors.toList());
    int number=0;
    for(int i=0;i<A.length;i++){
    if(Collections.frequency(list, A[i])==1){
    number=A[i];
    }
    }
    return number;
    }

  8. Have you unit tested it … this does not work …for eg int arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3}; output should be 2 …

  9. This is wat I came up with:

    public static int singleNumber(int[] numbers)
    {
    if(numbers == null)
    throw new NullPointerException(“Parameter numbers cannot be null.”);
    if(numbers.length == 0)
    throw new IllegalArgumentException(“Parameter numbers cannot be empty.”);

    Set oddNumbers = new HashSet();
    for(int i=0; i<numbers.length; i++) {
    int number = numbers[i];
    if(oddNumbers.contains(number))
    oddNumbers.remove(number);
    else
    oddNumbers.add(number);
    }

    if(oddNumbers.size() != 1)
    throw new IllegalArgumentException("Parameter number does not contain a single number.");

    return oddNumbers.iterator().next();
    }

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