From 9707fbc521b16e79a43ef588e4f4a77454a30fdf Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Wed, 2 Jun 2021 02:41:59 -0700
Subject: [PATCH] =?UTF-8?q?Update=200347.=E5=89=8DK=E4=B8=AA=E9=AB=98?=
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---
 problems/0347.前K个高频元素.md | 28 +++++++++++++++++++++++++++-
 1 file changed, 27 insertions(+), 1 deletion(-)

diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md
index 841584b4..71af618e 100644
--- a/problems/0347.前K个高频元素.md
+++ b/problems/0347.前K个高频元素.md
@@ -162,7 +162,33 @@ class Solution {
 
 
 Python：
-
+```python
+#时间复杂度：O(nlogk)
+#空间复杂度：O(n)
+import heapq
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        #要统计元素出现频率
+        map_ = {} #nums[i]:对应出现的次数
+        for i in range(len(nums)):
+            map_[nums[i]] = map_.get(nums[i], 0) + 1
+        
+        #对频率排序
+        #定义一个小顶堆，大小为k
+        pri_que = [] #小顶堆
+        
+        #用固定大小为k的小顶堆，扫面所有频率的数值
+        for key, freq in map_.items():
+            heapq.heappush(pri_que, (freq, key))
+            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
+                heapq.heappop(pri_que)
+        
+        #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
+        result = [0] * k
+        for i in range(k-1, -1, -1):
+            result[i] = heapq.heappop(pri_que)[1]
+        return result
+```
 
 Go：