189 - 347 前k个高频元素
题目
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]
说明:
你可以假设给定的 k 总是合理的,且 1 ≤ k ≤ 数组中不相同的元素的个数。
你的算法的时间复杂度必须优于 O(n log n) , n 是数组的大小。
解答
第一反应是用map,存一下所有数字出现的频率,然后输出最高的。。
看了题解发现,用hashmap存频率是无法避免的,关键是如何快速有效把高频元素取出来
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
from heapq import heappush
from heapq import heappop
hs = {}
for num in nums:
if num not in hs:
hs[num] = 1
else:
hs[num] += 1
heap = []
for i in hs:
heappush(heap, (-hs[i], i))
ans = []
for _ in range(k):
p = heappop(heap)
ans.append(p[1])
return ans
Runtime: 108 ms, faster than 76.04% of Python3 online submissions for Top K Frequent Elements.
Memory Usage: 17.4 MB, less than 6.25% of Python3 online submissions for Top K Frequent Elements.
python:我有一个方法。。。
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
from collections import Counter
most_common = Counter(nums).most_common()
fin = []
for l in range(0, k):
fin.append(most_common[l][0])
return fin
Runtime: 104 ms, faster than 88.26% of Python3 online submissions for Top K Frequent Elements.
Memory Usage: 17.2 MB, less than 6.25% of Python3 online submissions for Top K Frequent Elements.
还能更加简洁。。
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
from collections import Counter
from heapq import nlargest
count = Counter(nums)
return nlargest(k, count.keys(), key=count.get)
Runtime: 96 ms, faster than 98.80% of Python3 online submissions for Top K Frequent Elements.
Memory Usage: 17.1 MB, less than 6.25% of Python3 online submissions for Top K Frequent Elements.
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