[LeetCode] 215. array K-th largest element

Topic links: https://leetcode-cn.com/problems/kth-largest-element-in-an-array/ )

Subject description:

Find the unsorted array k largest elements. Please note that you need to find is the first k largest elements in the array is sorted, rather than the k distinct elements.

Example:

Example 1:

输入: [3,2,1,5,6,4] 和 k = 2
输出: 5

Example 2:

输入: [3,2,3,1,2,4,5,5,6] 和 k = 4
输出: 4

Description:

You may assume that k is always active, and 1 ≤ k ≤ length of the array.

Ideas:

Sort by title

Thinking a: \ (nlog (the n-) \) Sort

Two ideas: \ (nlog (k) \) heapsort

Library Functions
Handwriting heap row

Three ideas: The average time \ (O (n-) \) , the worst \ (O (n ^ 2) \) fast discharge portion

Each time a quick drain can be fixed number of positions, can be determined whether only looking at the position of!

Code:

A thought: Sort

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        return sorted(nums, reverse = True)[k - 1]

Thinking two: heapsort

1, library functions

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        return heapq.nlargest(k, nums)[-1]

Thinking two:

2, handwriting

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        
        
        def adjust_heap(idx, max_len):
            left = 2 * idx + 1
            right = 2 * idx + 2
            max_loc = idx
            if left < max_len and nums[max_loc] < nums[left]:
                max_loc = left
            if right < max_len and nums[max_loc] < nums[right]:
                max_loc = right
            if max_loc != idx:
                nums[idx], nums[max_loc] = nums[max_loc], nums[idx]
                adjust_heap(max_loc, max_len)
        
        # 建堆
        n = len(nums)
        for i in range(n // 2 - 1, -1, -1):
            adjust_heap(i, n)
        #print(nums)
        res = None
        for i in range(1, k + 1):
            #print(nums)
            res = nums[0]
            nums[0], nums[-i] = nums[-i], nums[0]
            adjust_heap(0, n - i)
        return res

Thinking three: fast row

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        def partition(left, right):
            pivot = nums[left]
            l = left + 1
            r = right
            while l <= r:
                if nums[l] < pivot and nums[r] > pivot:
                    nums[l], nums[r] = nums[r], nums[l]
                if nums[l] >= pivot:
                    l += 1
                if nums[r] <= pivot:
                    r -= 1
            nums[r], nums[left] = nums[left], nums[r]
            return r
        left = 0
        right = len(nums) - 1
        while 1:
            idx = partition(left, right)
            if idx == k - 1:
                return nums[idx]
            if idx < k - 1:
                left = idx + 1
            if idx > k - 1:
                right = idx - 1