leetcode Kth Largest Element in a Stream——要熟悉heapq使用

703. Kth Largest Element in a Stream

Easy

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.

Example:

int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3);   // returns 4
kthLargest.add(5);   // returns 5
kthLargest.add(10);  // returns 5
kthLargest.add(9);   // returns 8
kthLargest.add(4);   // returns 8

Note:

You may assume that nums‘ length ≥ k-1 and k ≥ 1.

"""
 1 import heapq
 2 def heapsort(iterable):
 3     h = []
 4     for i in iterable:
 5         heapq.heappush(h, i)
 6     return [heapq.heappop(h) for i in range(len(h))]
 7
 8 # method 1: sort to list
 9 s = [3, 5, 1, 2, 4, 6, 0, 1]
10 print(heapsort(s))
11 ‘‘‘
12 [0, 1, 1, 2, 3, 4, 5, 6]
13 ‘‘‘
"""
import heapq

class KthLargest(object):

    def __init__(self, k, nums):
        """
        :type k: int
        :type nums: List[int]
        """
        self.arr = []
        self.k = k
        for i in nums:
            if len(self.arr) < self.k:
                heapq.heappush(self.arr, i)
            else:
                if i > self.arr[0]:
                #if heapq.nsmallest(1, self.arr)[0] < i:
                    heapq.heapreplace(self.arr, i)
                    #heapq.heappop(self.arr)
                    #heapq.heappush(self.arr, i)

    def add(self, val):
        """
        :type val: int
        :rtype: int
        """
        if len(self.arr) < self.k:
            heapq.heappush(self.arr, val)
        else:
            if val > self.arr[0]:
            #if heapq.nsmallest(1, self.arr)[0] < val:
                heapq.heapreplace(self.arr, val)
                #heapq.heappop(self.arr)
                #heapq.heappush(self.arr, val)
        assert len(self.arr) == self.k
        return self.arr[0]
        #return heapq.nsmallest(1, self.arr)[0]

# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)

注意：如果使用heapq.nsmallest(1, self.arr)[0]来返回heap的最小值会有超时问题。

原文地址：https://www.cnblogs.com/bonelee/p/10090580.html