05 August 2008
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:
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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.
用优先队列实现一个最小堆，记录第k个，比较简单直接。
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class KthLargest {
PriorityQueue<Integer> pq;
int k;
public KthLargest(int k, int[] nums) {
pq = new PriorityQueue<>(k);
this.k = k;
for (int num : nums) {
add(num);
}
}
public int add(int val) {
if (pq.size() < k) {
pq.offer(val);
}else if (pq.peek() < val) {
pq.poll();
pq.offer(val);
}
return pq.peek();
}
}
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest obj = new KthLargest(k, nums);
* int param_1 = obj.add(val);
*/