문제
Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.
Example 1:
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Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:
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Input: nums = [1], k = 1
Output: [1]
Constraints:
1 <= nums.length <= 105104 <= nums[i] <= 104kis in the range[1, the number of unique elements in the array].- It is guaranteed that the answer is unique.
[풀이]
1. 단순 array 사용
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class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
numb = list(set(nums))
#print(numb)
lst = []
for i in range(len(numb)):
cnt = 0
cnt = nums.count(numb[i])
lst.append([numb[i],cnt])
new = sorted(lst, key=lambda x:x[1], reverse=True)
ans = []
for i in range(len(new)):
ans.append(new[i][0])
if len(ans) == k:
return ans
📌주목
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input: list = [[1,2],[2,3],[3,1]]
sorted(list, key=lambda x:x[1], reverse=True)
output: [[2,3],[1,2],[3,1]]
- key=lambda x : x[1], reverse=True ; 리스트 내의 1번째 인덱스를 기준으로 오름차순으로 정렬!
2. 다른 사람 풀이.. Dictionary를 사용한 방법!
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class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
frequency = {}
for num in nums:
if num not in frequency:
frequency[num] = 1
else:
frequency[num] = frequency[num] + 1
frequency = dict(sorted(frequency.items(), key=lambda x: x[1], reverse=True))
result = list(frequency.keys())[:k]
return result