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  1. Kth Largest Element in a Stream 难度 简单

https://leetcode-cn.com/problems/kth-largest-element-in-a-stream/

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) Appends the integer val to the stream and 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]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);// which means find the 3rd largest element in array [4, 5, 8, 2]
kthLargest.add(3);   // return 4. Now the sorted array is 8 5 4 3 2 the 3rd largest is 4.
kthLargest.add(5);   // return 5. Now the sorted array is 8 5 5 4 3 2. 
kthLargest.add(10);  // return 5. Now the sorted array is 10 8 5 4 3 2. 
kthLargest.add(9);   // return 8. Now the sorted array is 10 9 8 5 4 3 2. 
kthLargest.add(4);   // return 8. Now the sorted array is 10 9 8 5 4 4 3 2. 

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.

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相关标签

  • Tree
  • Design
  • Binary Search Tree
  • Binary Tree
  • Data Stream
  • Heap (Priority Queue)

相似题目

  • Kth Largest Element in an Array 中等

slow - Time: O(nlogn); Space O(1). because of sorting

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.k = k
        self.nums = nums

    def add(self, val: int) -> int:
        self.nums.append(val)
        self.nums = sorted(self.nums, reverse=True)
        return self.nums[self.k-1]

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

Partition - Time: O(n); Space: O(1)