<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 sortArray(self, nums: List[int]) -> List[int]:
        if sum(nums)==nums[0]*len(nums):return nums

        if sum(nums)==nums[1]*len(nums)+1:
            nums[0], nums[-1] = nums[-1], nums[0]
            return nums

        def quickSort(start, end):
            if start >= end:
                return -1

            flag = start - 1
            median = (start + end) // 2
            nums[end], nums[median] = nums[median], nums[end]
            for index in range(start, end + 1):
                if nums[index] <=  nums[end]:

                    flag += 1
                    if index > flag:
                        nums[index], nums[flag] = nums[flag], nums[index]

            # swap flag + 1, target
            pivot = flag
            quickSort(start, pivot - 1)
            quickSort(pivot + 1, end)


        quickSort(0, len(nums) - 1) #數字
        return nums


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