84. 柱状图中最大的矩形

• 思路
  • example
  • 暴力双指针（中心扩散）: . 或者固定左边界方法，同样复杂度。
  • 0 <= heights[i] <= 104
  • 单调栈
    • 栈递增 （list: 从左到右）

[2,1,5,6,2,3] extend to [0,2,1,5,6,2,3,0] （前后补0)
stack = [0] (实际stack存储元素下标）
stack = [0,2] heights[i] >= heights[stack[-1]]
stack = [0, 1] heights[i] < heights[stack[-1]], 处理以stack[-1]对应位置为高度的矩形面积，while 循环直到 条件不满足， 然后i位置入栈
stack = [0, 1, 5, 6]
stack = [0, 1, 2] 处理6，5高度
stack = [0, 1, 2, 3]
stack = [0, 0] 处理3，2，1高度

• 复杂度. 时间：, 空间:

另一个例子
[2,1,2], ans = 3

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        heights = [0] + heights + [0]
        n = len(heights)
        stack = [0]
        res = 0
        for i in range(n):
            if heights[i] >= heights[stack[-1]]:
                stack.append(i)
            else: # heights[i] < heights[stack[-1]]
                while heights[i] < heights[stack[-1]]: # stack 至少含有一个0
                    idx = stack.pop()
                    left = stack[-1] + 1 # !!!
                    res = max(res, (i-left) * heights[idx])
                stack.append(i)
        return res  

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        heights = [0] + heights + [0]
        n = len(heights) 
        stack = [0] 
        res = 0 
        for i in range(1, n):
            if heights[i] >= heights[stack[-1]]:
                stack.append(i)  
            else:
                while heights[i] < heights[stack[-1]]:
                    idx = stack.pop()
                    left = stack[-1] + 1 # !!!
                    res = max(res, heights[idx]*(i-left))  
                stack.append(i) 
        return res   

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        heights = [0] + heights + [0] 
        n = len(heights) 
        stack = [0]
        res = 0
        for i in range(1, n):
            height = heights[i] 
            while stack and height < heights[stack[-1]]:
                idx = stack.pop() 
                left = stack[-1] + 1 # !!! 以heights[idx]为高度的起始点. e.g., [2,1,2]
                res = max(res, heights[idx] * (i - left)) 
            stack.append(i)    
        return res