<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]:
        def quick_sort(arr, l, r): # 左闭右闭
            if l < r:  # 【Ensure that we have at least two elements to sort】
                pivot_index = partition(arr, l, r)
                quick_sort(arr, l, pivot_index)  # Sort the left half
                quick_sort(arr, pivot_index + 1, r)  # Sort the right half
            return arr # 如果不return 数组，稍微调一调也是对的

        def partition(arr, l, r):
            pivot = arr[(l + r) // 2]  # Choosing the middle element as pivot
            # 或： pivot = nums[random.randint(l, r)]
            while True:
                # while l < len(arr) and arr[l] < pivot:  # Move right until an element >= pivot is found
                while arr[l] < pivot:  
                    l += 1
                # while l < len(arr) and arr[r] > pivot:  # Move left until an element <= pivot is found
                while arr[r] > pivot:  
                    r -= 1
                if l >= r:  # 此时右指针一定是左半部分的结尾！# If pointers have crossed, partitioning is done
                # partition 函数中的判断条件 if l >- r: 中的运算符应为 >= 而不是 >-。

                    return r
                arr[l], arr[r] = arr[r], arr[l]  # Swap elements
                l += 1
                r -= 1
        # random.shuffle(nums)
        return quick_sort(nums, 0, len(nums) - 1) # 右边界应该是 len(nums) - 1 而不是 len(nums)，否则会引发数组越界错误。
    
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