在列表中找到第k个最大元素

Question

Given a list, find the k-th largest element in the list.

Input: list = [3, 5, 2, 4, 6, 8], k = 3

Output: 5

def findKthLargest(nums, k):
    pass

print(findKthLargest([3, 5, 2, 4, 6, 8], 3))
# 5 

Answer 1

I found two ways to solve this. First we would be sorting the array. So all we'd have to do is return the k-last index.

def findKthLargest1(nums, k):

    nums.sort()
    return nums[-k]

but there is a more interesting way to solve this problem, we can use Heaps. In general, when you hear about "smallest" or "largest". You should think: I need Heaps.

import heapq

def findKthLargest2(nums, k):

    minHeap = []
    heapq.heapify(minHeap)

    for x in nums:
        heapq.heappush(minHeap, x)
        if len(minHeap) > k:
            heapq.heappop(minHeap)

    return heapq.heappop(minHeap);

Answer 2

您还可以使用：

sorted(my_list)[-k]

Answer 3

sorted function sorts items in ascending order by default. You call also define reverse parameter (set True for descending order) and get the kth largest:

sorted(nums, reverse=True)[k-1]

Answer 4

try to do in time complexity < (n log n)

All solutions provided have n log n complexity.

The following code

def findKthLargest(nums, k):
    for _ in range(k):
        s = max(nums)
        t.remove(s)
    return s

will require 2(kn - k(k - 1)/2) operations, so its complexity is O(kn). If k is a constant (not a function of n), complexity is linear O(n). If k is a function of n but k < log n for all k, n, complexity is still < n log n. And if k = O(n) then complexity is O(n^2) which is > n log n and sorting will work faster.

Answer 5

k = int(input()) l = [6,2,1,9,5] l.sort() print(l[-k])