Sword finger offer II 040 Largest rectangle in matrix

The finger of the sword Offer II 040. The largest rectangle in the matrix

class Solution:
    def largestRectangleArea(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
    def maximalRectangle(self, matrix: List[str]) -> int:
        if not matrix:
            return 0
        row,col=len(matrix),len(matrix[0])
        height,ans=[0]*col,0
        for i in range(row):
            for j in range (col):
                if matrix[i][j]=='0': 
                    height[j]=0
                elif i>0 and matrix[i-1][j]=='0':   # The rest  and matrix[i][j]=1 and matrix[i-1][j]=0
                    height[j]=int(matrix[i][j])
                else:            # first line   And  matrix[i][j]=1 Or  The rest   And  matrix[i][j]=1  And  matrix[i-1][j]=1
                    height[j]+=1
            ans=max(ans,Solution.largestRectangleArea(height))   
        return ans