LeetCode-703. Kth Largest Element in a Stream

最新推荐文章于 2025-05-08 07:15:00 发布

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原文链接：http://www.cnblogs.com/zhacai/p/10566282.html

博客介绍了设计一个类来找出数据流中第k大元素的方法。可利用小顶堆（优先队列）解决，时间复杂度为O(logk)；也可使用二分查找找到数据保存位置，添加后返回第k个值，时间复杂度为O(n)。

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.

利用小顶堆解决，即优先队列，时间复杂度为O(logk)

 1 class KthLargest {//优先队列 堆 mytip
 2 
 3     PriorityQueue<Integer> heap;
 4     int k;
 5     
 6     public KthLargest(int k, int[] nums) {
 7         heap = new PriorityQueue<Integer>(k);
 8         this.k =k;
 9         for (int x:nums) {
10             this.add(x);
11         }
12         
13     }
14     
15     public int add(int val) {
16         if(heap.size()<k){
17             heap.add(val);
18         }
19         else if(val>heap.peek()){
20             heap.poll();
21             heap.add(val);
22         }
23         return heap.peek();
24     }
25 }

使用通常二分查找，找到数据需保存的位置，添加后返回第k个值，二分查找时间复杂度为O(logn)（可优化为O(logk)），ArrayList指定位置插入时间复杂度为O(n)，故时间复杂度为O(n)。

 1 class KthLargest {//二分查找 my
 2 
 3     List<Integer> topk ;
 4     int k;
 5     
 6     public KthLargest(int k, int[] nums) {
 7         topk= new ArrayList<Integer>();
 8         this.k =k;
 9         for (int x:nums) {
10             this.add(x);
11         }
12         
13     }
14     
15     public int add(int val) {
16         int pos= this.getPos(val,0,topk.size());
17         this.topk.add(pos,val);
18         if(topk.size()<=this.k){
19             return topk.get(topk.size()-1);
20         }
21         else{
22             return topk.get(k-1);
23         }
24     }
25     private int getPos(int val,int start,int end){
26         if(start>=end){
27             return start;
28         }
29         int mid =(start+end)/2;
30         if(this.topk.get(mid)<val){
31             return getPos(val,start,mid);
32         }
33         else {
34             return getPos(val,mid+1,end);
35         }
36     }
37 }