260. 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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

这里有两个不同的数只出现过1次，一路异或过去，出现两次的数异或结果变成0，剩下的数是两个target之间有差异的位，换句话说一个数是0，另一个数的对应位就是1。用diff &= -diff能得到最后一位1。然后用这个特殊位分离开两个数。举个例子：

二进制数，正负数互换，则要把全部数码（包括符号位）都求反加1。
＋5：0101
－5：1010+1 = 1011

+5 & -5 = 0001

代码如下：

public class Solution {
    public int[] singleNumber(int[] nums) {
        int diff = 0;
        for (int num: nums) {
            diff ^= num;
        }
        diff &= -diff;
        int target1 = 0, target2 = 0;
        for (int num: nums) {
            if ((num & diff) == 0) {
                target1 ^= num;
            } else {
                target2 ^= num;
            }
        }
        return new int[]{target1, target2};
    }
}