mycode 71.43%
class Solution(object): def topKFrequent(self, nums, k): """ :type nums: List[int] :type k: int :rtype: List[int] """ if not nums: return [] from collections import Counter s = Counter(nums).most_common() res = [] for val,count in s: res.append(val) if len(res) == k: return res
参考：
思路：
heapq--该模块提供了堆排序算法的实现。堆是二叉树，最大堆中父节点大于或等于两个子节点，最小堆父节点小于或等于两个子节点。
如果需要获取堆中最大或最小的范围值，则可以使用heapq.nlargest() 或heapq.nsmallest() 函数
下面例子中heapq.nlargest的第二个参数遍历，每一个值代入第三个参数中，得到最终用来排序的数组成的列表
import heapq from pprint import pprint portfolio = [ {‘name‘: ‘IBM‘, ‘shares‘: 100, ‘price‘: 91.1}, {‘name‘: ‘AAPL‘, ‘shares‘: 50, ‘price‘: 543.22}, {‘name‘: ‘FB‘, ‘shares‘: 200, ‘price‘: 21.09}, {‘name‘: ‘HPQ‘, ‘shares‘: 35, ‘price‘: 31.75}, {‘name‘: ‘YHOO‘, ‘shares‘: 45, ‘price‘: 16.35}, {‘name‘: ‘ACME‘, ‘shares‘: 75, ‘price‘: 115.65} ] cheap = heapq.nsmallest(3, portfolio, key=lambda s: s[‘price‘]) expensive = heapq.nlargest(3, portfolio, key=lambda s: s[‘price‘]) pprint(cheap) pprint(expensive) """ 输出： [{‘name‘: ‘YHOO‘, ‘price‘: 16.35, ‘shares‘: 45}, {‘name‘: ‘FB‘, ‘price‘: 21.09, ‘shares‘: 200}, {‘name‘: ‘HPQ‘, ‘price‘: 31.75, ‘shares‘: 35}] [{‘name‘: ‘AAPL‘, ‘price‘: 543.22, ‘shares‘: 50}, {‘name‘: ‘ACME‘, ‘price‘: 115.65, ‘shares‘: 75}, {‘name‘: ‘IBM‘, ‘price‘: 91.1, ‘shares‘: 100}]
遍历每个key，作为第三个参数的参数，得到对应的值，他们组合了用来排序的所有值
import heapq class Solution(object): def topKFrequent(self, nums, k): """ :type nums: List[int] :type k: int :rtype: List[int] """ if not k: return 0 mydict={} for i in nums: if i in mydict: mydict[i]+=1 else: mydict[i]=0 return heapq.nlargest(k,mydict.keys(),mydict.get)
leetcode-mid-sorting and searching -347. Top K Frequent Elements
原文：https://www.cnblogs.com/rosyYY/p/10975270.html