Sword finger offer II 039 Histogram maximum rectangular area monotonic stack

The finger of the sword Offer II 039. Maximum rectangular area of histogram
The width of the matrix must be , Start from both sides of the column at the top of the stack until you meet the width of the column that is higher than the column .
Because the column height stored in the monotonic stack is increasing , So the column in front of the column on the top of the stack must be shorter than the column on the top of the stack , Similarly, the currently scanned column is also shorter than the column at the top of the stack , So the width of the highest matrix with the top column as the top is determined , Then the area is determined .

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        heights.append(0)   # The height of the last column is 0, All the remaining elements of the list will pop Calculate the maximum rectangular area when it is height 
        lenh=len(heights)
        maxArea=0
        h=[-1]
        for i in range (lenh):
            while h[-1]!=-1 and heights[i]<heights[h[-1]]:
                area_height=heights[h[-1]]
                h.pop()
                area_width=i-h[-1]-1     # When there is only one column element left , Left boundary index It can be regarded as -1
                maxArea=max(maxArea,area_height*area_width)
            h.append(i)        # here heights[i]>heights[h[-1]]
        return maxArea