[Leetcode]703. Kth Largest Element in a Stream 数据流中的第 K 大元素
作者:互联网
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 integerk
and the stream of integersnums
.int add(int val)
Returns the element representing thekth
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]); kthLargest.add(3); // return 4 kthLargest.add(5); // return 5 kthLargest.add(10); // return 5 kthLargest.add(9); // return 8 kthLargest.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 toadd
. - It is guaranteed that there will be at least
k
elements in the array when you search for thekth
element.
题意
设计一个找到数据流中第 k
大元素的类(class)。注意是排序后的第 k
大元素,不是第 k
个不同的元素。
思路
- 先试着对给定的an integer k and an integer array nums进行一下构造函数constructoer的设计使得其包含数据流最初这些给定参数
- 每次从数据流新加入的data ,就跟堆顶元素比较一下:比堆顶元素小,直接忽略;比堆顶元素大,就堆顶元素让位。而该堆mantain了k size个最大元素,所以每次返回的堆顶元素,就是第K 大元素
代码
/* Time:O(n * logk) 假设数据个数为n对每个数据,最坏情况下,每次都打掉堆顶元素,priorityqueue内部又得重新排序,ksize的priorityqueue的排序时间为logk, 所以总时间是 n 乘以 logk Space: O(k) 额外开了 k size的priorityqueue做辅助 */ class KthLargest { PriorityQueue<Integer> minHeap; int k; //constructor public KthLargest (int k, int[] nums) { this.k = k; minHeap = new PriorityQueue<>(k); // Java PriorityQueue is a min heap in default for(int n : nums){ add(n); } } public int add(int val) { if(minHeap.size() < k){ minHeap.offer(val); }else if(minHeap.peek() < val){ minHeap.poll(); minHeap.offer(val); } return minHeap.peek(); } }
标签:kthLargest,元素,Stream,int,KthLargest,Element,add,Kth,minHeap 来源: https://www.cnblogs.com/liuliu5151/p/14204554.html