题目

思路

1. The first step is to build a hash map. Python provides us both a dictionary structure for the hash map and a method Counter in the collections library to build the hash map we need.This step takes O(N) time where N is number of elements in the list.
2. The second step is to build a heap. The time complexity of adding an element in a heap is O(log(k)) and we do it N times that means O(Nlog(k)) time complexity for this step.
3. The last step to build an output list has O(klog(k)) time complexity.

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)

代码

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.