動機

當初想的太複雜了，還想用兩個heap

Problem

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.

Implement KthLargest class:

  • KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
  • int add(int val) Returns the element representing the kth largest element in the stream.

 

Example 1:

Input[KthLargest, add, add, add, add, add][[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]Output[null, 4, 5, 5, 8, 8]ExplanationKthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);kthLargest.add(3);   // return 4kthLargest.add(5);   // return 5kthLargest.add(10);  // return 5kthLargest.add(9);   // return 8kthLargest.add(4);   // return 8

 

Constraints:

  • 1 <= k <= 104
  • 0 <= nums.length <= 104
  • -104 <= nums[i] <= 104
  • -104 <= val <= 104
  • At most 104 calls will be made to add.
  • It is guaranteed that there will be at least k elements in the array when you search for the kth element.

Sol

Min heap保持最多k個

class KthLargest:
    def __init__(self, k: int, nums: List[int]):
        self.minh = nums
        self.k = k
        heapify(self.minh)
        while len(self.minh) > k:
            heappop(self.minh)

    def add(self, val: int) -> int:
        heappush(self.minh,val)
        #print(self.minh, "=>")
        if len(self.minh) > self.k:
            heappop(self.minh)
        #print(self.minh)
        return self.minh[0]