题目描述；

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.

思考：

找出第k大的数，那就是第N-k+1小的数了，采用快速排序，时间复杂程度为O(nlogN)

代码：

public class Solution {

	public static int findKthLargest(int[] nums, int k) {

		int n = nums.length;
		int site = n - k + 1;

		if(n == 0){
			return 0;
		}


		return find(nums, site - 1, 0, n - 1);
	}


	public static int find(int[] nums, int site, int left, int right){
		int res = 0;

		if(left <= right){
			int base = nums[left];
			int l = left;
			int r = right;

			while(l < r){
				while(nums[r] >= base && l < r){

					r--;
				}
				if(l < r){
					nums[l] = nums[r];
					l++;	
				}
				while(nums[l] <= base && l < r){
					l++;
				}
				if(l < r){
					nums[r] = nums[l];
					r--;	
				}		
			}

			nums[r] = base;
			if(r == site){
				res = base;
			}

			else if(r> site){
				int left1 = left;
				int right1 = r -1;
				res = find(nums, site, left1, right1);
			}		
			else{
				int left2 = r + 1;
				int right2 = right;
				res = find(nums, site, left2, right2);
			}
		}

		return res;
	}
}