215. Kth Largest Element in an Array

public class Solution {
    private int sort(int[] a, int lo, int hi, int k) {
        int p = partition(a, lo, hi);
        if (k == p) {
            return a[p];
        } else if (k < p) {
            return sort(a, lo, p - 1, k);
        } else {
            return sort(a, p + 1, hi, k);
        }
    }

    // This is an in-place implementation of partition in Algorithm (4th Edition)
    private int partition(int[] a, int lo, int hi) {
        int i = lo;
        int j = hi + 1;
        int v = a[lo];
        while (true) {
            while (i < hi && a[++i] < v) {}
            while (j > lo && a[--j] > v) {}
            if (i >= j) {
                break;
            }
            exch(a, i, j);
        }
        exch(a, lo, j);
        return j;
    }

    // Exchanges the elements in a[i] and a[j]
    private void exch(int[] a, int i, int j) {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }

    // Fisher–Yates shuffle algorithm
    private void shuffle(int[] a) {
        Random random = new Random();
        for (int i = a.length - 1; i >= 0; i--) {
            // 0 <= r <= i
            int r = random.nextInt(i + 1);
            exch(a, i, r);
        }
    }

    // If we sort the array in ascending order, 
    // the kth element of an array will be the kth smallest element.
    // To find the kth largest element, we can pass k = length(Array) – k.
    public int findKthLargest(int[] a, int k) {
        shuffle(a);
        return sort(a, 0, a.length - 1, a.length - k);
    }
}