Wiggle Sort
<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 = [3,5,2,1,6,4]
<strong>Output:</strong> [3,5,1,6,2,4]
<strong>Explanation:</strong> [1,6,2,5,3,4] is also accepted.
</pre>

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

<pre>
<strong>Input:</strong> nums = [6,6,5,6,3,8]
<strong>Output:</strong> [6,6,5,6,3,8]
</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;= 10<sup>4</sup></code></li>
	<li>It is guaranteed that there will be an answer for the given input <code>nums</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve the problem in <code>O(n)</code> time complexity?</p>
Greedy; Array; Sorting
class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        for i in range(1, len(nums)):
            if (i % 2 == 1 and nums[i] < nums[i - 1]) or (
                i % 2 == 0 and nums[i] > nums[i - 1]
            ):
                nums[i], nums[i - 1] = nums[i - 1], nums[i]
