<p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>

<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>

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

<pre>
<strong>Input:</strong> nums = [5,2,3,1]
<strong>Output:</strong> [1,2,3,5]
<strong>Explanation:</strong> After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,1,1,2,0,0]
<strong>Output:</strong> [0,0,1,1,2,5]
<strong>Explanation:</strong> Note that the values of nums are not necessairly unique.
</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>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution:
    # def quickSort(self, nums):
    #     if len(nums) <= 1:
    #         return nums

    #     pivot = nums[0]
    #     left = [num for num in nums[1:] if num < pivot]
    #     right = [num for num in nums[1:] if num >= pivot]
    #     return self.quickSort(left) + [pivot] + self.quickSort(right)

    def merge(self, l1, l2):
        i1, i2 = 0, 0
        res = []

        while i1 < len(l1) and i2 < len(l2):
            if l1[i1] < l2[i2]:
                res.append(l1[i1])
                i1 += 1
            else:
                res.append(l2[i2])
                i2 += 1

        if i1 < len(l1):
            res += l1[i1:]
        if i2 < len(l2):
            res += l2[i2:]
        return res

    def sortArray(self, nums: List[int]) -> List[int]:
        lists = [[num] for num in nums]
        while len(lists) > 1:
            newLists = []
            for i in range(0, len(lists), 2):
                if i + 1 < len(lists):
                    newLists.append(self.merge(lists[i], lists[i + 1]))
                else:
                    newLists.append(lists[i])
            lists = newLists

        return lists[0]
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