LeetCode *** 260. Single Number III

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本文深入解析了一种高效的算法，用于找出数组中仅出现一次的两个元素，而其他元素均出现两次。该算法巧妙地利用了位运算，确保运行时复杂度为O(n)，且仅使用常数级额外空间。通过实例分析，读者可以理解其核心思想并掌握其实现细节。

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题目：

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:

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

分析：

看的题解，不得不说很绝妙。good。

代码：

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        vector<int>res={0,0};
        
        int d=0;
        for(int num:nums)d^=num;
        d&=-d;
        for(int num:nums)if(num&d)res[0]^=num;
        else res[1]^=num;
        
        return res;
        
    }
};