Single Number 解答

Question

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

Solution 1 -- Set

We can use a hash set to record each integer's appearing time. Time complexity O(n), space cost O(n)

public class Solution {
    public int singleNumber(int[] nums) {
        Set<Integer> counts = new HashSet<Integer>();
        int length = nums.length, result = nums[0];
        for (int i = 0; i < length; i++) {
            int tmp = nums[i];
            if (counts.contains(tmp))
                counts.remove(tmp);
            else
                counts.add(tmp);
        }
        for (int tmp : counts)
            result = tmp;
        return result;
    }
}

Solution 2 -- Bit Manipulation

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;
}