Update 0084.柱状图中最大的矩形.md 动态规划和单调栈 Python3 版本

problems/0084.柱状图中最大的矩形.md

@@ -191,4 +191,57 @@ public:
 
 这里我依然建议大家按部就班把版本一写出来，把情况一二三分析清楚，然后在精简代码到版本二。 直接看版本二容易忽略细节！
 
+## 其他语言版本 
+
+Java: 
+
+Python:
+
+动态规划
+```python3
+class Solution:
+    def largestRectangleArea(self, heights: List[int]) -> int:
+        result = 0
+        minleftindex, minrightindex = [0]*len(heights), [0]*len(heights)
+        
+        minleftindex[0]=-1
+        for i in range(1,len(heights)):
+            t = i-1
+            while t>=0 and heights[t]>=heights[i]: t=minleftindex[t]
+            minleftindex[i]=t
+            
+        minrightindex[-1]=len(heights)
+        for i in range(len(heights)-2,-1,-1):
+            t=i+1
+            while t<len(heights) and heights[t]>=heights[i]: t=minrightindex[t]
+            minrightindex[i]=t
+        
+        for i in range(0,len(heights)):
+            left = minleftindex[i]
+            right = minrightindex[i]
+            summ = (right-left-1)*heights[i]
+            result = max(result,summ)
+        return result
+```
+单调栈 版本二
+```python3
+class Solution:
+    def largestRectangleArea(self, heights: List[int]) -> int:
+        heights.insert(0,0) # 数组头部加入元素0
+        heights.append(0) # 数组尾部加入元素0
+        st = [0]
+        result = 0
+        for i in range(1,len(heights)):
+            while st!=[] and heights[i]<heights[st[-1]]:
+                midh = heights[st[-1]]
+                st.pop()
+                if st!=[]:
+                    minrightindex = i
+                    minleftindex = st[-1]
+                    summ = (minrightindex-minleftindex-1)*midh
+                    result = max(summ,result)
+            st.append(i)
+        return result
+```
+
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>