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.

Example:

Input:  [1,2,1,3,2,5]
Output: [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?

先把所有數xor一遍得到一個由兩個只出現一次的數xor後得到的結果，取最右邊的bit來當作分別兩個只出現一次的數的標準，再loop一次所有數把數分成兩堆去做xor。

// Two pass method with XOR Bit Manipulation
vector<int> singleNumber(vector<int>& nums) { // time: O(n); space: O(1)
    int diff = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());
    diff &= -diff; // -diff = ~diff + 1
    vector<int> res(2, 0);
    for (int num : nums) {
        if ((num & diff) == 0) res[0] ^= num;
        else res[1] ^= num;
    }
    return res;
}