LeetCode Single Number III

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:

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

本题初看可以使用hashmap，我的代码如下：

public class Solution {
    public int[] singleNumber(int[] nums) {
        HashMap hashmap = new HashMap();
        for ( int i =0 ;i<nums.length;i++) {
            if (hashmap.containsKey(nums[i])) hashmap.remove(nums[i]);
              else hashmap.put(nums[i],i);
        }
        Iterator it = hashmap.keySet().iterator();
        int ans[] = new int[2];
        int x = 0;
        while (it.hasNext()) {
            ans[x] = (int)it.next();
            x++;
        }
        return ans;
    }
}

但在经过看discuss后，发现可以使用XOR(^)来进行计算

如果一组数里面有两个单个数，那么XOR的结果一定是这两个数的XOR，这个结果中的每个二进制下的1都是两个数不同的地方，而其他数会在XOR运算中抵消掉。那么可以针对第一个结果中的1进行分类，多余的两个不同的数一定有一个是1而另一个是0，又因为其他数都会抵消掉，那么可以对这个位进行判断后进行XOR。