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Update 0347.前K个高频元素.md
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@ -162,7 +162,33 @@ class Solution {
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Python:
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Python:
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```python
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#时间复杂度:O(nlogk)
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#空间复杂度:O(n)
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import heapq
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class Solution:
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def topKFrequent(self, nums: List[int], k: int) -> List[int]:
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#要统计元素出现频率
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map_ = {} #nums[i]:对应出现的次数
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for i in range(len(nums)):
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map_[nums[i]] = map_.get(nums[i], 0) + 1
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#对频率排序
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#定义一个小顶堆,大小为k
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pri_que = [] #小顶堆
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#用固定大小为k的小顶堆,扫面所有频率的数值
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for key, freq in map_.items():
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heapq.heappush(pri_que, (freq, key))
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if len(pri_que) > k: #如果堆的大小大于了K,则队列弹出,保证堆的大小一直为k
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heapq.heappop(pri_que)
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#找出前K个高频元素,因为小顶堆先弹出的是最小的,所以倒叙来输出到数组
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result = [0] * k
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for i in range(k-1, -1, -1):
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result[i] = heapq.heappop(pri_que)[1]
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return result
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```
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Go:
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Go:
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