leecode题号347：前K个高频元素，python实现_python前k个高频元素 给定一个非空的整数数组,返回其中出现频率前k高的元素。例如

给定一个非空的整数数组，返回其中出现频率前 k 高的元素。

示例 1:

输入: nums = [1,1,1,2,2,3], k = 2
输出: [1,2]
示例 2:

输入: nums = [1], k = 1
输出: [1]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/top-k-frequent-elements
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解法一：字典排序

class Solution(object):
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        dict = {}
        for i in nums:
            if i not in dict:
                dict[i] = 1
            else:
                dict[i] += 1
        print('dict1:')
        print(dict)
        dict_sort = sorted(dict.items(), key = lambda x: x[1],reverse = True)
        print('sorted dict:')
        print(dict_sort)
        res = []
        for i in range(k):
            res.append(dict_sort[i][0])
        return res
            
        
    

解法二：桶排序

class Solution(object):
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        maps = {}
        #生成字典映射
        for i in nums:
            # print('i:%d,maps.get(i,0):%d' %(i, maps.get(i,0)))
            maps[i] = maps.get(i,0) + 1
        max_time = max(maps.values())
        print(maps)
        #根据最大次数生成桶
        TongList = [[] for i in range(max_time+1)]
        for key ,value in maps.items():
            TongList[value].append(key)
            print(TongList)
        res = []
        for i in range(max_time, 0, -1):
            print('i:%d' %(i))
            if TongList[i]:
                res.extend(TongList[i])
            if len(res) >= k:
                return res[:k]