<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:
        """
        Do not return anything, modify nums in-place instead.
        """
        
        nums_sorted = self.__mergeSort(nums)

        n = len(nums)

        left = nums_sorted[:n//2+n%2]

        right = nums_sorted[n//2+n%2:]

        i = len(left) - 1

        j = len(right) - 1

        for index in range(n):

            if index % 2 == 0:

                nums[index] = left[i]

                i -= 1

            else:

                nums[index] = right[j]

                j -= 1


    def __mergeSort(self,nums):

        if len(nums) < 2:

            return nums

        left = self.__mergeSort(nums[:len(nums)//2])

        right = self.__mergeSort(nums[len(nums)//2:])

        new_nums = []

        m = len(left)

        n = len(right)

        i = 0

        j = 0

        while i < m and j < n:

            if left[i] < right[j]:

                new_nums.append(left[i])

                i += 1

            else:

                new_nums.append(right[j])

                j += 1

        while i < m:

            new_nums.append(left[i])

            i += 1

        while j < n:

            new_nums.append(right[j])

            j += 1

        return new_nums
        
19800