【Leetcode/Problem Solving Report】 215. Kth Largest Element in an Array

    Given an unsorted array, find the kth largest number in it.

    Using the idea of ​​divide and conquer, take a random number p from the array each time, put all the numbers greater than p on the left of p, and put the rest on the right of p, and mark the final subscript of p in the array as i, Then p is the i+1th largest number of the array.

    If i+1 == k, the k-th largest number is p, and the algorithm ends.

    If i+1 < k, the kth largest number is on the right side of p, and continue to find the k - i - 1th largest number in the array to the right of p.

    If i + 1 > k, the kth largest number is on the left of p, and continue to find the kth largest number in the left array of p.

    In the first version, I opened an extra array to store the numbers larger than p and smaller than p. This code is concise and easy to understand, but it will exceed the space limit.

class Solution(object):
	def findKthLargest(self, nums, k):
		"""
		:type nums: List[int]
		:type k: int
		:rtype: int
		"""
		pivot = nums[0]
		small = [i for i in nums[1:] if i <= pivot]
		large = [i for i in nums[1:] if i > pivot]
		if len(large) + 1 == k: return pivot
		elif len(large) + 1 < k: return self.findKthLargest(small, k - len(large) - 1)
		else: return self.findKthLargest(large, k)

    Instead of using the extra space, just modify the original array.

class Solution(object):
	def findKthLargest(self, nums, k):
		"""
		:type nums: List[int]
		:type k: int
		:rtype: int
		"""
		return self.findK(nums, 0, k, len(nums))

	def findK(self, nums, start, k, end):
		pivot = nums[start]
		last_big = start # index of the last number greater than p
		for i in range(start + 1, end):
			if nums[i] > pivot:
				last_big += 1
				nums[last_big], nums[i] = nums[i], nums[last_big]
		nums[last_big], nums[start] = nums[start], nums[last_big]      # last_big变为p的下标
		if last_big + 1 == k: return pivot
		elif last_big + 1 > k: return self.findK(nums, start, k, last_big)
		else: return self.findK(nums, last_big + 1, k, end) # Note that k does not need to be changed