Given the array nums consisting of 2n elements in the form [x1,x2,...,xn,y1,y2,...,yn].
Return the array in the form [x1,y1,x2,y2,...,xn,yn].
Example 1:

Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7] 
Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].
Example 2:

Input: nums = [1,2,3,4,4,3,2,1], n = 4
Output: [1,4,2,3,3,2,4,1]
Example 3:

Input: nums = [1,1,2,2], n = 2
Output: [1,2,1,2]
 

Constraints:

1 <= n <= 500
nums.length == 2n
1 <= nums[i] <= 10^3

非常常规的解法

class Solution {
public:
    vector<int> shuffle(vector<int>& nums, int n) {
        vector<int> res;
        int elem_cnt = nums.size();
        for(int i = 0;i < elem_cnt / 2 ;i++)
        {
            res.push_back(nums[i]);
            res.push_back(nums[i + n]);
        }
        
        return res;
    }
};

一般人想不到的解法，看了评论区大神的解法拍案叫绝！这里主要利用了vector中元素最大是1000这样一条特性。
第一个循环将nums前半部分保存在后半部分
第二个循环将元素保存在最终位置，由于都是位运算，方法二性能要明显由于方法一，而且这里的空间复杂度是O(1)

class Solution {
public:
    vector<int> shuffle(vector<int>& nums, int n) {
        for (int i = 0; i < n; ++i) {
            nums[i + n] |= (nums[i] << 10);
        }
        
        for (int i = 0; i < n; ++i) {
            nums[i * 2] = (nums[i + n] & 0xFFC00) >> 10;     // 11111111110000000000 == 0xFFC00
            nums[i * 2 + 1] = nums[i + n] & 0x003FF;        // 00000000001111111111 == 0x003FF
        }
        
        return nums;
    }
};