Kth Largest Element in an Array
<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>largest element in the array</em>.</p>

<p>Note that it is the <code>k<sup>th</sup></code> largest element in the sorted order, not the <code>k<sup>th</sup></code> distinct element.</p>

<p>Can you solve it without sorting?</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [3,2,1,5,6,4], k = 2
<strong>Output:</strong> 5
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [3,2,3,1,2,4,5,5,6], k = 4
<strong>Output:</strong> 4
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Divide and Conquer; Quickselect; Sorting; Heap (Priority Queue)
class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        int n = nums.size();
        k = n - k;
        auto quickSort = [&](auto&& quickSort, int l, int r) -> int {
            if (l == r) {
                return nums[l];
            }
            int i = l - 1, j = r + 1;
            int x = nums[(l + r) >> 1];
            while (i < j) {
                while (nums[++i] < x) {
                }
                while (nums[--j] > x) {
                }
                if (i < j) {
                    swap(nums[i], nums[j]);
                }
            }
            if (j < k) {
                return quickSort(quickSort, j + 1, r);
            }
            return quickSort(quickSort, l, j);
        };
        return quickSort(quickSort, 0, n - 1);
    }
};
