Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.For example,
Given [3,2,1,5,6,4] and k = 2, return 5.Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.

 1 public class Solution {
 2     public int findKthLargest(int[] nums, int k) {
 3         int len = nums.length;
 4         return findKthSmallest(nums, 0, len-1, len-k+1);
 5     }
 6
 7     public int findKthSmallest(int[] nums, int start, int end, int k) {
 8         int l = start;
 9         int r = end;
10         int pivot = end;
11         while (l < r) {
12             while (l<r && nums[l] < nums[pivot]) {
13                 l++;
14             }
15             while (l<r && nums[r] >= nums[pivot]) {
16                 r--;
17             }
18             if (l == r) break;
19             swap(nums, l, r);
20         }
21         swap(nums, l, pivot);
22         if (l+1 == k) return nums[l];
23         else if (l+1 < k) return findKthSmallest(nums, l+1, end, k);
24         else return findKthSmallest(nums, start, l-1, k);
25     }
26
27     public void swap(int[] nums, int l, int r) {
28         int temp = nums[l];
29         nums[l] = nums[r];
30         nums[r] = temp;
31     }
32 }

 1 public class Solution {
 2     public int findKthLargest(int[] nums, int k) {
 3         PriorityQueue<Integer> p = new PriorityQueue<Integer>();
 4         for(int i = 0 ; i < nums.length; i++){
 5             p.add(nums[i]);
 6             if(p.size()>k) p.poll();
 7         }
 8         return p.poll();
 9     }
10 }