From 9319ce9cdcab03027644ce1cff57b4060d53bc99 Mon Sep 17 00:00:00 2001
From: cherrypicker <cherrypicker404@gmail.com>
Date: Sat, 23 Dec 2023 08:34:38 +0800
Subject: [PATCH] =?UTF-8?q?Update0020.=E6=9C=89=E6=95=88=E7=9A=84=E6=8B=AC?=
 =?UTF-8?q?=E5=8F=B7=20Go=E7=A4=BA=E4=BE=8B=E4=BB=A3=E7=A0=81?=
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原有的实现已经足够简洁清晰，这个版本的代码增加了注释，提升了可读性。
---
 problems/0020.有效的括号.md | 48 ++++++++++++++++++++++----------
 1 file changed, 33 insertions(+), 15 deletions(-)

diff --git a/problems/0020.有效的括号.md b/problems/0020.有效的括号.md
index 045c79ee..17fbe2be 100644
--- a/problems/0020.有效的括号.md
+++ b/problems/0020.有效的括号.md
@@ -218,23 +218,41 @@ class Solution:
 ### Go：
 
 ```Go
+// 思路： 使用栈来进行括号的匹配
+// 时间复杂度 O(n)
+// 空间复杂度 O(n)
 func isValid(s string) bool {
-    hash := map[byte]byte{')':'(', ']':'[', '}':'{'}
-    stack := make([]byte, 0)
-    if s == "" {
-        return true
-    }
+	// 使用切片模拟栈的行为
+	stack := make([]rune, 0)
 
-    for i := 0; i < len(s); i++ {
-        if s[i] == '(' || s[i] == '[' || s[i] == '{' {
-            stack = append(stack, s[i])
-        } else if len(stack) > 0 && stack[len(stack)-1] == hash[s[i]] {
-            stack = stack[:len(stack)-1]
-        } else {
-            return false
-        }
-    }
-    return len(stack) == 0
+	// m 用于记录某个右括号对应的左括号
+	m := make(map[rune]rune)
+	m[')'] = '('
+	m[']'] = '['
+	m['}'] = '{'
+
+	// 遍历字符串中的 rune
+	for _, c := range s {
+		// 左括号直接入栈
+		if c == '(' || c == '[' || c == '{' {
+			stack = append(stack, c)
+		} else {
+			// 如果是右括号，先判断栈内是否还有元素
+			if len(stack) == 0 {
+				return false
+			}
+			// 再判断栈顶元素是否能够匹配
+			peek := stack[len(stack)-1]
+			if peek != m[c] {
+				return false
+			}
+			// 模拟栈顶弹出
+			stack = stack[:len(stack)-1]
+		}
+	}
+
+	// 若栈中不再包含元素，则能完全匹配
+	return len(stack) == 0
 }
 ```