703. 数据流中的第 K 大元素
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) 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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/kth-largest-element-in-a-stream
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
排序
class KthLargest: def __init__(self, k: int, nums: List[int]): self.nums=nums self.k=k # self.nums.sort(reverse=True) # while len(self.nums)>k: # self.nums.pop() def add(self, val: int) -> int: self.nums.append(val) self.nums.sort(reverse=True) # if len(self.nums)>self.k: # self.nums.pop() return self.nums[self.k-1] # Your KthLargest object will be instantiated and called as such: # obj = KthLargest(k, nums) # param_1 = obj.add(val)
class KthLargest: def __init__(self, k: int, nums: List[int]): self.nums=nums self.k=k self.nums.sort(reverse=True) while len(self.nums)>k: self.nums.pop() def add(self, val: int) -> int: self.nums.append(val) self.nums.sort(reverse=True) if len(self.nums)>self.k: self.nums.pop() return self.nums[-1] # Your KthLargest object will be instantiated and called as such: # obj = KthLargest(k, nums) # param_1 = obj.add(val)
堆
class KthLargest: def __init__(self, k: int, nums: List[int]): self.nums=nums self.k=k heapq.heapify(self.nums) while len(self.nums)>k: heapq.heappop(self.nums) def add(self, val: int) -> int: if len(self.nums)<self.k: heapq.heappush(self.nums,val) elif val>self.nums[0]: heapq.heapreplace(self.nums,val) return self.nums[0] # Your KthLargest object will be instantiated and called as such: # obj = KthLargest(k, nums) # param_1 = obj.add(val)