Question

剑指 Offer II 059. 数据流的第 K 大数值

题目描述

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/

解法

方法一

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 any) {
  *h = append(*h, x.(int))
}
func (h *IntHeap) Pop() any {
  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);
 */