在未排序的数组中找到第 k 个最大的元素。请注意，你需要找的是数组排序后的第 k 个最大的元素，而不是第 k 个不同的元素。

输入: [3,2,1,5,6,4] 和 k = 2
输出: 5

输入: [3,2,3,1,2,4,5,5,6] 和 k = 4
输出: 4

说明:
你可以假设 k 总是有效的，且 1 ≤ k ≤ 数组的长度。

class Solution(object):
    def quick_sort(self, array, l, r):
        if l < r:
            q = self.partition(array, l, r)
            self.quick_sort(array, l, q - 1)
            self.quick_sort(array, q + 1, r)
 
    def partition(self, array, l, r):
        x = array[r]
        i = l - 1
        for j in range(l, r):
            if array[j] <= x:
                i += 1
                array[i], array[j] = array[j], array[i]
        array[i + 1], array[r] = array[r], array[i+1]
        return i + 1
        
    
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        if not nums or k > len(nums):
            return nums
        self.quick_sort(nums, 0, len(nums)-1)
        return nums[-k]

优化版本：

class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        if not nums or k > len(nums):
            return nums
        
        l = 0
        r = len(nums) - 1
        while True:
            index = self.partition(nums, l, r)
            print(nums)
            if index == k - 1:
                return nums[index]
            elif index > k - 1:
                r = index - 1
            else:
                l = index + 1
                
    def partition(self, nums, left, right):
        pivot = nums[left]
        l = left
        r = right
        while l <= r:
            while l <= r and nums[l] >= pivot:
                l += 1
            while l <= r and nums[r] < pivot:
                r -= 1
            print(nums[l], nums[r])
            if l <= r and nums[l] < pivot and nums[r] >= pivot:
                nums[l], nums[r] = nums[r], nums[l]
                l += 1
                r -= 1
        nums[left], nums[r] = nums[r], nums[left]
        return