Update 0084.柱状图中最大的矩形.md

补全所有python解法，补充comment

Author: casnz1601

problems/0084.柱状图中最大的矩形.md

@@ -281,52 +281,137 @@ class Solution {
 
 Python:
 
-动态规划
 ```python3
+
+# 双指针；暴力解法（leetcode超时）
 class Solution:
     def largestRectangleArea(self, heights: List[int]) -> int:
+        # 从左向右遍历：以每一根柱子为主心骨（当前轮最高的参照物），迭代直到找到左侧和右侧各第一个矮一级的柱子
+        res = 0
+
+        for i in range(len(heights)):
+            left = i
+            right = i
+            # 向左侧遍历：寻找第一个矮一级的柱子
+            for _ in range(left, -1, -1):
+                if heights[left] < heights[i]:
+                    break
+                left -= 1
+            # 向右侧遍历：寻找第一个矮一级的柱子
+            for _ in range(right, len(heights)):
+                if heights[right] < heights[i]:
+                    break
+                right += 1
+                
+            width = right - left - 1
+            height = heights[i]
+            res = max(res, width * height)
+
+        return res
+
+# DP动态规划
+class Solution:
+    def largestRectangleArea(self, heights: List[int]) -> int:
+        size = len(heights)
+        # 两个DP数列储存的均是下标index
+        min_left_index = [0] * size
+        min_right_index = [0] * size
         result = 0
-        minleftindex, minrightindex = [0]*len(heights), [0]*len(heights)
+
+        # 记录每个柱子的左侧第一个矮一级的柱子的下标
+        min_left_index[0] = -1  # 初始化防止while死循环
+        for i in range(1, size):
+            # 以当前柱子为主心骨，向左迭代寻找次级柱子
+            temp = i - 1
+            while temp >= 0 and heights[temp] >= heights[i]:
+                # 当左侧的柱子持续较高时，尝试这个高柱子自己的次级柱子（DP
+                temp = min_left_index[temp]
+            # 当找到左侧矮一级的目标柱子时
+            min_left_index[i] = temp
 
-        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
+        # 记录每个柱子的右侧第一个矮一级的柱子的下标
+        min_right_index[size-1] = size  # 初始化防止while死循环
+        for i in range(size-2, -1, -1):
+            # 以当前柱子为主心骨，向右迭代寻找次级柱子
+            temp = i + 1
+            while temp < size and heights[temp] >= heights[i]:
+                # 当右侧的柱子持续较高时，尝试这个高柱子自己的次级柱子（DP
+                temp = min_right_index[temp]
+            # 当找到右侧矮一级的目标柱子时
+            min_right_index[i] = temp
+        
+        for i in range(size):
+            area = heights[i] * (min_right_index[i] - min_left_index[i] - 1)
+            result = max(area, result)
 
-        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
+
+# 单调栈
 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]
+        # Monotonic Stack
+        '''
+        找每个柱子左右侧的第一个高度值小于该柱子的柱子
+        单调栈：栈顶到栈底：从大到小（每插入一个新的小数值时，都要弹出先前的大数值）
+        栈顶，栈顶的下一个元素，即将入栈的元素：这三个元素组成了最大面积的高度和宽度
+        情况一：当前遍历的元素heights[i]大于栈顶元素的情况
+        情况二：当前遍历的元素heights[i]等于栈顶元素的情况
+        情况三：当前遍历的元素heights[i]小于栈顶元素的情况
+        '''
+
+        # 输入数组首尾各补上一个0（与42.接雨水不同的是，本题原首尾的两个柱子可以作为核心柱进行最大面积尝试
+        heights.insert(0, 0)
+        heights.append(0)
+        stack = [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)
+        for i in range(1, len(heights)):
+            # 情况一
+            if heights[i] > heights[stack[-1]]:
+                stack.append(i)
+            # 情况二
+            elif heights[i] == heights[stack[-1]]:
+                stack.pop()
+                stack.append(i)
+            # 情况三
+            else:
+                # 抛出所有较高的柱子
+                while stack and heights[i] < heights[stack[-1]]:
+                    # 栈顶就是中间的柱子，主心骨
+                    mid_index = stack[-1]
+                    stack.pop()
+                    if stack:
+                        left_index = stack[-1]
+                        right_index = i
+                        width = right_index - left_index - 1
+                        height = heights[mid_index]
+                        result = max(result, width * height)
+                stack.append(i)
         return result
+
+# 单调栈精简
+class Solution:
+    def largestRectangleArea(self, heights: List[int]) -> int:
+        heights.insert(0, 0)
+        heights.append(0)
+        stack = [0]
+        result = 0
+        for i in range(1, len(heights)):
+            while stack and heights[i] < heights[stack[-1]]:
+                mid_height = heights[stack[-1]]
+                stack.pop()
+                if stack:
+                    # area = width * height
+                    area = (i - stack[-1] - 1) * mid_height
+                    result = max(area, result)
+            stack.append(i)
+        return result
+
+
+
+
+
 ```
+*****
 
 JavaScript:
 ```javascript