215. 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.

java

使用可快排的方式，但是应该还有更好地解决方法，稍后更新

class Solution {    public int findKthLargest(int[] nums, int k) {        if (nums == null || nums.length == 0 || k < 0) {            return -1;        }        if (k > nums.length) {            return -1;        }        quickSort(nums, 0, nums.length - 1);        return nums[nums.length - k];    }    private void quickSort(int[] nums, int start, int end) {        if (start >= end) {            return;        }        int left = start;        int right = end;        int pivot = nums[(left + right) / 2];        while (left <= right) {            while (left <= right && nums[left] < pivot) {                left++;            }            while (left <= right && nums[right] > pivot) {                right--;            }            if (left <= right) {                int temp = nums[left];                nums[left] = nums[right];                nums[right] = temp;                left++;                right--;            }        }        quickSort(nums, start, right);        quickSort(nums, left, end);    }}

python

class Solution(object):    def findKthLargest(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: int        """        if nums == None or len(nums) == 0 or k < 0:            return -1        if k > len(nums):            return -1        self.quickSort(nums, 0, len(nums) - 1)        return nums[len(nums) - k]            def quickSort(self, nums, start, end):        if start >= end:            return        left, right = start, end        pivot = nums[(left + right) / 2]        while left <= right:            while left <= right and nums[left] < pivot:                left += 1            while left <= right and nums[right] > pivot:                right -= 1            if left <= right:                nums[left], nums[right] = nums[right], nums[left]                left += 1                right -= 1        self.quickSort(nums, start, right)        self.quickSort(nums, left, end)