<p>Given an integer array <code>nums</code>, reorder it such that <code>nums[0] &lt; nums[1] &gt; nums[2] &lt; nums[3]...</code>.</p>

<p>You may assume the input array always has a valid answer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,1,1,6,4]
<strong>Output:</strong> [1,6,1,5,1,4]
<strong>Explanation:</strong> [1,4,1,5,1,6] is also accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,2,2,3,1]
<strong>Output:</strong> [2,3,1,3,1,2]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 5000</code></li>
	<li>It is guaranteed that there will be an answer for the given input <code>nums</code>.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow Up:</strong> Can you do it in <code>O(n)</code> time and/or <strong>in-place</strong> with <code>O(1)</code> extra space?
class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        auto n = nums.size();

        auto mid = nums.begin() + (n >> 1);
        std::nth_element(nums.begin(), mid, nums.end());

        int median = *mid;

        for (int i = 0, j = n - 1, k = 0; k <= j;) {
            auto rk = R(k, n), rkv = nums[rk];
            if (rkv > median) {
                if (i < k)
                    std::swap(nums[rk], nums[R(i, n)]);
                ++i, ++k;
            } else if (rkv < median) {
                std::swap(nums[rk], nums[R(j--, n)]);
            } else {
                ++k;
            }
        }
    }

private:
    static constexpr int R(int p, int n) {
        return ((p << 1) | 1) % (n | 1);
    }
};
19600