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.

思路

利用快速排序的思想进行分治。在一趟排序过程中，如果关键数（pivot）右侧有k-1个数，即说明恰好有k-1个数大于此关键数（pivot），从而找到第k大的数，递归结束；否则就对关键数两侧的部分分别进行递归即可。

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class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        if (len(nums) == 1):
            return nums[0]
        else:
            pivot = nums[0]
            left = 0
            right = len(nums) - 1
            mid = 0
            while (left < right):
                while (right >= 0 and left < right):
                    if (nums[right] < pivot):
                        nums[right], nums[mid] = nums[mid], nums[right]
                        mid = right
                        break
                    right -= 1
                if(left >= right):
                    break
                while (left < len(nums) and left < right):
                    if (nums[left] > pivot):
                        nums[left], nums[mid] = nums[mid], nums[left]
                        mid = left
                        break
                    left += 1
            if (len(nums) - mid == k):
                return nums[mid]
            elif (len(nums) - mid > k):
                return self.findKthLargest(nums[mid + 1:], k)
            else:
                return self.findKthLargest(nums[:mid], k - (len(nums) - mid) )