LeetCode-python 85.最大矩形

题目链接
难度：困难       类型：动态规划

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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

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矩形的面积等于，分别找出和即可

1. 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
2. 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

本文链接：https://www.jianshu.com/p/67cd5ba8802c