From 571defa00789aa9764abb6aa7efe54944ed42a49 Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Fri, 20 Aug 2021 02:15:03 +0800
Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200020.=E6=9C=89=E6=95=88?=
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补充了题主的方法，改写了原先冗余的做法，可读性提升，PEP8标准
---
 problems/0020.有效的括号.md | 44 +++++++++++++++++++++++++-------
 1 file changed, 35 insertions(+), 9 deletions(-)

diff --git a/problems/0020.有效的括号.md b/problems/0020.有效的括号.md
index a26aa308..57729c38 100644
--- a/problems/0020.有效的括号.md
+++ b/problems/0020.有效的括号.md
@@ -162,18 +162,44 @@ class Solution {
 
 Python：
 ```python3
+# 方法一，仅使用栈，更省空间
 class Solution:
     def isValid(self, s: str) -> bool:
-        stack = []  # 保存还未匹配的左括号
-        mapping = {")": "(", "]": "[", "}": "{"}
-        for i in s:
-            if i in "([{":  # 当前是左括号，则入栈
-                stack.append(i)
-            elif stack and stack[-1] == mapping[i]:  # 当前是配对的右括号则出栈
-                stack.pop()
-            else:  # 不是匹配的右括号或者没有左括号与之匹配，则返回false
+        stack = []
+        
+        for item in s:
+            if item == '(':
+                stack.append(')')
+            elif item == '[':
+                stack.append(']')
+            elif item == '{':
+                stack.append('}')
+            elif not stack or stack[-1] != item:
                 return False
-        return stack == []  # 最后必须正好把左括号匹配完
+            else:
+                stack.pop()
+        
+        return True if not stack else False
+```
+
+```python3
+# 方法二，使用字典
+class Solution:
+    def isValid(self, s: str) -> bool:
+        stack = []
+        mapping = {
+            '(': ')',
+            '[': ']',
+            '{': '}'
+        }
+        for item in s:
+            if item in mapping.keys():
+                stack.append(mapping[item])
+            elif not stack or stack[-1] != item: 
+                return False
+            else: 
+                stack.pop()
+        return True if not stack else False
 ```
 
 Go：