From fbeb80f14ad127e1d47c549c1c2063715896a6bf Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Mon, 16 Aug 2021 18:24:43 +0800
Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200202.=E5=BF=AB=E4=B9=90?=
 =?UTF-8?q?=E6=95=B0.md=20python=E4=BB=A3=E7=A0=81=E7=AE=80=E5=8C=96?=
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利用闭包的特性，修改了原本python代码中不符合PEP3的写法，以及赘余的代码。思路保持一致
---
 ...2.\345\277\253\344\271\220\346\225\260.md" | 36 ++++++++++---------
 1 file changed, 20 insertions(+), 16 deletions(-)

diff --git "a/problems/0202.\345\277\253\344\271\220\346\225\260.md" "b/problems/0202.\345\277\253\344\271\220\346\225\260.md"
index 1c630b6a8e..2e784e6ac8 100644
--- "a/problems/0202.\345\277\253\344\271\220\346\225\260.md"
+++ "b/problems/0202.\345\277\253\344\271\220\346\225\260.md"
@@ -111,25 +111,29 @@ Python：
 ```python
 class Solution:
     def isHappy(self, n: int) -> bool:
-        set_ = set()
-        while 1:
-            sum_ = self.getSum(n)
-            if sum_ == 1:
+        def calculate_happy(num):
+            sum_ = 0
+            
+            # 从个位开始依次取，平方求和
+            while num:
+                sum_ += (num % 10) ** 2
+                num = num // 10
+            return sum_
+
+        # 记录中间结果
+        record = set()
+
+        while True:
+            n = calculate_happy(n)
+            if n == 1:
                 return True
-            #如果这个sum曾经出现过，说明已经陷入了无限循环了，立刻return false
-            if sum_ in set_:
+            
+            # 如果中间结果重复出现，说明陷入死循环了，该数不是快乐数
+            if n in record:
                 return False
             else:
-                set_.add(sum_)
-            n = sum_
-            
-    #取数值各个位上的单数之和
-    def getSum(self, n):
-        sum_ = 0
-        while n > 0:
-            sum_ += (n%10) * (n%10)
-            n //= 10
-        return sum_
+                record.add(n)
+
 ```
 
 Go：