2024-03-14 18:24:40
In Python there is a method nlargest in heapq library which has the same O(klog(k)) time complexity and combines two last steps in one line.
优先队列也就是最大最小堆。时间复杂度为O(nlogk)
class Solution:
def topKFrequent(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
count = collections.Counter(nums)
return heapq.nlargest(k, count.keys(), key=count.get)
heapq.nlargest(n, iterable[, key])
从迭代器对象iterable中返回前n个最大的元素列表,其中关键字参数key用于匹配是字典对象的iterable,即选用字典中哪个key作为比较字段。
使用get方法获取其计数.