From 14223a2fa6c4a93d9e782a6b4e32e73682c5e051 Mon Sep 17 00:00:00 2001
From: Yuan Yuan <yyccphil@gmail.com>
Date: Wed, 4 Dec 2024 12:33:40 -0600
Subject: [PATCH] =?UTF-8?q?fix:=20Update=201365=EF=BC=8C=E5=BB=BA=E8=AE=AE?=
 =?UTF-8?q?=E4=B8=8D=E4=BD=BF=E7=94=A8built-in=20func=E4=BD=9C=E4=B8=BAvar?=
 =?UTF-8?q?=20name?=
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题解中使用`hash`作为变量名，而hash本身也是python3的built-in函数，故建议更改变量名
---
 problems/1365.有多少小于当前数字的数字.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/problems/1365.有多少小于当前数字的数字.md b/problems/1365.有多少小于当前数字的数字.md
index 22dd3226..95a270ff 100644
--- a/problems/1365.有多少小于当前数字的数字.md
+++ b/problems/1365.有多少小于当前数字的数字.md
@@ -144,13 +144,13 @@ public int[] smallerNumbersThanCurrent(int[] nums) {
 class Solution:
     def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
         res = nums[:]
-        hash = dict()
+        hash_dict = dict()
         res.sort() # 从小到大排序之后，元素下标就是小于当前数字的数字
         for i, num in enumerate(res):
-            if num  not in hash.keys(): # 遇到了相同的数字，那么不需要更新该 number 的情况
-                hash[num] = i       
+            if num  not in hash_dict.keys(): # 遇到了相同的数字，那么不需要更新该 number 的情况
+                hash_dict[num] = i       
         for i, num in enumerate(nums):
-            res[i] = hash[num]
+            res[i] = hash_dict[num]
         return res
 ```