<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) {
        std::sort(nums.begin(), nums.end());
        const int n = nums.size();
        std::vector<int> temp(std::next(nums.begin(), (n+1)/2), nums.end());
        for (int i=n-1; i>=0; --i) {
            if (i%2 == 0) {
                nums[i] = nums[i/2];
            } else {
                nums[i] = temp[(i-1) / 2];
            }
        }
        int k = 2;
        for (k=2; k<n; k+=2) {
            if (nums[k] == nums[k-1]) {
                break;
            }
        }
        if (k < n) {
            std::vector<int> tmp(nums.begin(), std::next(nums.begin(), k));
            for (int i=k; i<n; ++i) {
                nums[i-k] = nums[i];
            }
            for (int i=n-k; i<n; ++i) {
                nums[i] = tmp[i-n+k];
            }
        }
    }
};
20600