Sort an Array
<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 necessarily 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>
Array; Divide and Conquer; Bucket Sort; Counting Sort; Radix Sort; Sorting; Heap (Priority Queue); Merge Sort
class Solution {
    fun sortArray(nums: IntArray): IntArray {
        fun quickSort(left: Int, right: Int) {
            if (left >= right) {
                return
            }
            var i = left - 1
            var j = right + 1
            val pivot = nums[left]
            while (i < j) {
                while (nums[++i] < pivot) ;
                while (nums[--j] > pivot) ;
                if (i < j) {
                    val temp = nums[i]
                    nums[i] = nums[j]
                    nums[j] = temp
                }
            }
            quickSort(left, j)
            quickSort(j + 1, right)
        }
        quickSort(0, nums.size - 1)
        return nums
    }
}
