目录

题目描述

解法1:堆

解法2:快排

题目描述

Given an integer array nums and an integer k, return the kth largest element in the array.

Note that it is the kth largest element in the sorted order, not the kth distinct element.

You must solve it in O(n) time complexity.

Example 1:

Input: nums = [3,2,1,5,6,4], k = 2
Output: 5

Example 2:

Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
Output: 4

Constraints:

• 1 <= k <= nums.length <= 105
• -104 <= nums[i] <= 104

解法1:堆

维护一个大小为K的小顶堆

python:

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        import heapq as hq
        heap = nums[:k]
        hq.heapify(heap)
        for t in nums[k:]:
            if t > heap[0]:
                hq.heapreplace(heap, t)
        return heap[0]

c++：

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        priority_queue<int> q(nums.begin(), nums.end());
        for (int i = 0; i < k - 1; i++) {
            q.pop();
        }
        return q.top();
    }
};

解法2:快排

快排思想是根据某值将数组分成两部分，左边的均小于该数，右边的均大于该数

这里可以按照从大到小排序，找到应该放在第k-1位置的数即可

python：

class Solution:
    def partition(self, nums, l, r, k):
        if l >= r:
            return
        cur = nums[l]
        i = l
        j = r
        while l < r:
            while l < r and cur > nums[r]:
                r -= 1
            if l < r:
                nums[l] = nums[r]
                l += 1
            while l < r and nums[l] > cur:
                l += 1
            if l < r:
                nums[r] = nums[l]
                r -= 1
        nums[l] = cur
        if l == k-1:
            return
        elif l < k-1:
            self.partition(nums, l + 1, j, k)
        else:
            self.partition(nums, i, l - 1, k)
        

    def findKthLargest(self, nums: List[int], k: int) -> int:
        self.partition(nums, 0, len(nums) - 1, k)
        return nums[k-1]