From 40b392588699b33a750101ab63adbb3b6f4f6860 Mon Sep 17 00:00:00 2001
From: OrangeLiuSSE <143183385+OrangeLiuSSE@users.noreply.github.com>
Date: Thu, 7 Mar 2024 10:45:09 +0800
Subject: [PATCH] =?UTF-8?q?Update=200347.=E5=89=8DK=E4=B8=AA=E9=AB=98?=
 =?UTF-8?q?=E9=A2=91=E5=85=83=E7=B4=A0.md?=
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---
 problems/0347.前K个高频元素.md | 27 ++++++++++++++++++++++++++-
 1 file changed, 26 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 93d605f5..b340e185 100644
--- a/problems/0347.前K个高频元素.md
+++ b/problems/0347.前K个高频元素.md
@@ -218,7 +218,7 @@ class Solution {
 ```
 
 ### Python：
-
+解法一：
 ```python
 #时间复杂度：O(nlogk)
 #空间复杂度：O(n)
@@ -246,6 +246,31 @@ class Solution:
             result[i] = heapq.heappop(pri_que)[1]
         return result
 ```
+解法二：
+```python
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        # 使用字典统计数字出现次数
+        time_dict = defaultdict(int)
+        for num in nums:
+            time_dict[num] += 1
+        # 更改字典，key为出现次数，value为相应的数字的集合
+        index_dict = defaultdict(list)
+        for key in time_dict:
+            index_dict[time_dict[key]].append(key)
+        # 排序
+        key = list(index_dict.keys())
+        key.sort()
+        result = []
+        cnt = 0
+        # 获取前k项
+        while key and cnt != k:
+            result += index_dict[key[-1]]
+            cnt += len(index_dict[key[-1]])
+            key.pop()
+
+        return result[0: k]
+```
 
 ### Go：