From 4671287dd3494c3b2d7592450e1c7c92f13d0504 Mon Sep 17 00:00:00 2001
From: jinzhongjia <jinzhongjia@qq.com>
Date: Thu, 13 Oct 2022 00:34:44 +0800
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---
 problems/0084.柱状图中最大的矩形.md | 39 +++++++++++++++++++-
 1 file changed, 38 insertions(+), 1 deletion(-)

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 e085e455..144b4027 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
@@ -409,7 +409,44 @@ class Solution:
 
 
 ```
-*****
+
+Go:
+
+> 单调栈
+
+```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
+}
+
+```
 
 JavaScript:
 ```javascript