LeeCode-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?

题目很简单，但是通过这个题目让我见识了LeetCode众多大牛的存在。

代码中，第一次了使用accumulate函数，然后还要bit_xor<int>()之类的C++11的function object。

此处如果diff = INT_MIN, 那么 -diff 恰好会溢出到 INT_MIN. 之前并没有注意过这个问题。

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        int diff = accumulate(begin(nums), end(nums), 0, bit_xor<int>());
        diff &= -diff;
        vector<int> result = {0, 0};
        for (auto num: nums) {
            result[!(diff & num)] ^= num;
        }
        return result;
    }
};