Easy
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]);
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
104
calls will be made to add
.k
elements in the array when you search for the kth
element.import java.util.PriorityQueue
class KthLargest(private val maxSize: Int, nums: IntArray) {
private val heap: PriorityQueue<Int> = PriorityQueue()
init {
for (num in nums) {
add(num)
}
}
fun add(`val`: Int): Int {
if (heap.size < maxSize) {
heap.add(`val`)
} else if (heap.peek() < `val`) {
heap.add(`val`)
heap.poll()
}
return heap.peek()
}
}
/*
* Your KthLargest object will be instantiated and called as such:
* var obj = KthLargest(k, nums)
* var param_1 = obj.add(`val`)
*/