问题描述
Given a non-empty array of integers, return the k most frequent elements.
Example 1:
Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:
Input: nums = [1], k = 1
Output: [1]
Note:
You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
思路
loop一遍nums
,把每个数字和对应的出现次数存进一个dict
把dict
的key
和value
位置对换。 注意,避免把对换后的数据存在另一个dictionary中,因为不同的数字可能出现的次数是一样的,会被覆盖
之后根据频次排序,将频次排在前k个的对应数字存进一个数组并返回即可
def topKFrequent(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
dict = {}
for i in nums:
if i in dict:
dict[i] += 1
else:
dict[i] = 1
revDict = []
for j in dict.keys():
revDict.append([dict.get(j), j])
sortByH = sorted(revDict, key=lambda s: s[0])
ans = []
while k and sortByH:
ans.append(sortByH.pop()[1])
k -= 1
return ans
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