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。
请实现 KthLargest
类:
KthLargest(int k, int[] nums)
使用整数k
和整数流nums
初始化对象。int add(int val)
将val
插入数据流nums
后,返回当前数据流中第k
大的元素。
示例:
输入: ["KthLargest", "add", "add", "add", "add", "add"] [[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]] 输出: [null, 4, 5, 5, 8, 8] 解释: 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
提示:
1 <= k <= 104
0 <= nums.length <= 104
-104 <= nums[i] <= 104
-104 <= val <= 104
- 最多调用
add
方法104
次 - 题目数据保证,在查找第
k
大元素时,数组中至少有k
个元素
注意:本题与主站 703 题相同: https://leetcode.cn/problems/kth-largest-element-in-a-stream/
小根堆存放最大的 k 个元素,那么堆顶就是第 k 大的元素。
class KthLargest:
def __init__(self, k: int, nums: List[int]):
self.q = []
self.size = k
for num in nums:
self.add(num)
def add(self, val: int) -> int:
heappush(self.q, val)
if len(self.q) > self.size:
heappop(self.q)
return self.q[0]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)
class KthLargest {
private PriorityQueue<Integer> q;
private int size;
public KthLargest(int k, int[] nums) {
q = new PriorityQueue<>(k);
size = k;
for (int num : nums) {
add(num);
}
}
public int add(int val) {
q.offer(val);
if (q.size() > size) {
q.poll();
}
return q.peek();
}
}
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest obj = new KthLargest(k, nums);
* int param_1 = obj.add(val);
*/
class KthLargest {
public:
priority_queue<int, vector<int>, greater<int>> q;
int size;
KthLargest(int k, vector<int>& nums) {
size = k;
for (int num : nums) add(num);
}
int add(int val) {
q.push(val);
if (q.size() > size) q.pop();
return q.top();
}
};
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest* obj = new KthLargest(k, nums);
* int param_1 = obj->add(val);
*/
type KthLargest struct {
h *IntHeap
k int
}
func Constructor(k int, nums []int) KthLargest {
h := &IntHeap{}
heap.Init(h)
for _, v := range nums {
heap.Push(h, v)
}
for h.Len() > k {
heap.Pop(h)
}
return KthLargest{
h: h,
k: k,
}
}
func (this *KthLargest) Add(val int) int {
heap.Push(this.h, val)
for this.h.Len() > this.k {
heap.Pop(this.h)
}
return this.h.Top()
}
func connectSticks(sticks []int) int {
h := IntHeap(sticks)
heap.Init(&h)
res := 0
for h.Len() > 1 {
val := heap.Pop(&h).(int)
val += heap.Pop(&h).(int)
res += val
heap.Push(&h, val)
}
return res
}
type IntHeap []int
func (h IntHeap) Len() int { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h IntHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *IntHeap) Push(x interface{}) {
*h = append(*h, x.(int))
}
func (h *IntHeap) Pop() interface{} {
old := *h
n := len(old)
x := old[n-1]
*h = old[0 : n-1]
return x
}
func (h *IntHeap) Top() int {
if (*h).Len() == 0 {
return 0
}
return (*h)[0]
}
/**
* Your KthLargest object will be instantiated and called as such:
* obj := Constructor(k, nums);
* param_1 := obj.Add(val);
*/