Kth Largest Element in an Array

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.

Tips:

两种方法，heap的复杂度是O(nlogk)，quick sort的平均是O(n)，经过shuffle input几乎可以保证是O(n)。

还有Quick Select(D & C):

n+(n/2)+(n/4)..1 = n + (n-1) = O(2n-1) = O(n), because n/2+n/4+n/8+..1=n-1+

Code：

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        shuffle(nums);
        int start = 0, end = nums.length - 1;
        k = nums.length - k;
        while (start < end) {
            int pivot = partition(nums, start, end);
            if (pivot == k) return nums[pivot];
            else if (pivot < k) start = pivot + 1;
            else end = pivot - 1;
        }
        return nums[end];
    }

    private int partition(int[] nums, int start, int end) {
        int pivot = start;
        while (start <= end) {
            while (start <= end && nums[start] <= nums[pivot]) start++;
            while (start <= end && nums[end] >= nums[pivot]) end--;
            if (start < end) swap(nums, start, end);
        }
        swap(nums, pivot, end);
        return end;
    }

    private void shuffle(int[] nums) {
        Random random = new Random();
        for (int i = 0; i < nums.length; i++) {
            int key = random.nextInt(i + 1);
            swap(nums, i, key);
        }
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}