leetcode最大矩形_LeetCode-python 85.最大矩形

题目链接

难度：困难       类型：动态规划

给定一个仅包含 0 和 1 的二维二进制矩阵，找出只包含 1 的最大矩形，并返回其面积。

示例

输入:

[["1","0","1","0","0"],

["1","0","1","1","1"],

["1","1","1","1","1"],

["1","0","0","1","0"]]

输出: 6

解题思路

矩形的面积等于

，分别找出

和

即可

计算

row[i]是matrix的第i行，每行有n列，例如示例中，n=5

到达第

行时

当row[i][j] == '0' 时(其中

)，

当row[i][j] == '1' 时(其中

)，

例如，i = 2时，h = [3, 1, 3, 2, 2]，每一行都要求出h，一共求4组

找到

对于第2行的h，[3, 1, 3, 2, 2]，前三行所构成的矩形的面积可以是1，2，3，4，6，这是因为去了不同的

所造成的

当

是，面积等于6

[["x","x","x","x","x"],

["x","x","1","1","1"],

["x","x","1","1","1"],

["x","x","x","x","x"]

当

是，面积等于5

[["x","x","x","x","x"],

["x","x","x","x","x"],

["1","1","1","1","1"],

["x","x","x","x","x"]

找出所有的

和

的组合，像极了84题：柱状图中最大的矩形

代码实现

class Solution:

def maximalRectangle(self, matrix: List[List[str]]) -> int:

if not matrix or not matrix[0]:

return 0

n = len(matrix[0])

height = [0] * (n+1)

max_area = 0

for row in matrix:

# 计算h

for i in range(n):

height[i] = height[i]+1 if row[i]=='1' else 0

# 找出所有h和w的组合

stack = [-1]

for j in range(n + 1):

while height[j] < height[stack[-1]]:

h = height[stack.pop()]

w = j - 1 - stack[-1]

max_area = max(max_area, h * w)

stack.append(j)

return max_area