From c26fddc702a333a3ed36fa10ebbf0f71d617ce7b Mon Sep 17 00:00:00 2001
From: ShuangmingMa <52561813+ShuangmingMa@users.noreply.github.com>
Date: Sat, 16 Sep 2023 22:33:02 +0800
Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200084.=E6=9F=B1=E7=8A=B6?=
 =?UTF-8?q?=E5=9B=BE=E4=B8=AD=E6=9C=80=E5=A4=A7=E7=9A=84=E7=9F=A9=E5=BD=A2?=
 =?UTF-8?q?=20Go=E5=8D=95=E8=B0=83=E6=A0=88=E4=BB=A3=E7=A0=81=E7=9A=84?=
 =?UTF-8?q?=E6=A0=BC=E5=BC=8F?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0084.柱状图中最大的矩形.md | 56 ++++++++++----------
 1 file changed, 27 insertions(+), 29 deletions(-)

diff --git a/problems/0084.柱状图中最大的矩形.md b/problems/0084.柱状图中最大的矩形.md
index b54429ed..97dab4f3 100644
--- a/problems/0084.柱状图中最大的矩形.md
+++ b/problems/0084.柱状图中最大的矩形.md
@@ -478,36 +478,34 @@ class Solution:
 
 ```go
 func largestRectangleArea(heights []int) int {
- // 声明max并初始化为0
- max := 0
- // 使用切片实现栈
- stack := make([]int, 0)
- // 数组头部加入0
- heights = append([]int{0}, heights...)
- // 数组尾部加入0
- heights = append(heights, 0)
- // 初始化栈，序号从0开始
- stack = append(stack, 0)
- for i := 1; i < len(heights); i++ {
-  // 结束循环条件为：当即将入栈元素>top元素，也就是形成非单调递增的趋势
-  for heights[stack[len(stack)-1]] > heights[i] {
-   // mid 是top
-   mid := stack[len(stack)-1]
-   // 出栈
-   stack = stack[0 : len(stack)-1]
-   // left是top的下一位元素，i是将要入栈的元素
-   left := stack[len(stack)-1]
-   // 高度x宽度
-   tmp := heights[mid] * (i - left - 1)
-   if tmp > max {
-    max = tmp
-   }
-  }
-  stack = append(stack, i)
- }
- return max
+    max := 0
+    // 使用切片实现栈
+    stack := make([]int, 0)
+    // 数组头部加入0
+    heights = append([]int{0}, heights...)
+    // 数组尾部加入0
+    heights = append(heights, 0)
+    // 初始化栈，序号从0开始
+    stack = append(stack, 0)
+    for i := 1; i < len(heights); i ++ {
+        // 结束循环条件为：当即将入栈元素>top元素，也就是形成非单调递增的趋势
+        for heights[stack[len(stack) - 1]] > heights[i] {
+            // mid 是top
+            mid := stack[len(stack) - 1]
+            // 出栈
+            stack = stack[0 : len(stack) - 1]
+            // left是top的下一位元素，i是将要入栈的元素
+            left := stack[len(stack) - 1]
+            // 高度x宽度
+            tmp := heights[mid] * (i - left - 1)
+            if tmp > max {
+                max = tmp
+            }
+        }
+        stack = append(stack, i)
+    }
+    return max
 }
-
 ```
 
 ### JavaScript: