<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:
    def wiggleSort(self, nums: List[int]) -> None:
        nums.sort()
        size = len(nums)
        middle = size // 2
        is_pair = size % 2 == 0

        p0 = middle - 1 if is_pair else middle
        p1 = size - 1

        ans = []
        while len(ans) != size:
            ans.append(nums[p0])
            p0 -= 1

            if len(ans) == size:
                break

            ans.append(nums[p1])
            p1 -= 1
            
        for i in range(size):
            nums[i] = ans[i]








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