LeetCode开心刷题三十二天——85. Maximal Rectangle

85. Maximal Rectangle

Hard

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Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.

Example:

Input:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
Output: 6

IDEA:
https://www.twblogs.net/a/5bc0fa7d2b717711c9241544/zh-cn
The concept of monotone stack(单调栈) was introduced
ONE IMPORTANT THING:
dp[i][j] means the number of "1" in i rows before j columns.
dp[1][2]代表在第一行中的第二列之前1的个数。

#include
#include
#include<string>
#include
#include<set>
#include

 

 python version:
1.python version why don't duplicate the C++'s thought,this
method looks more complex.
2.this version complex to understand but I add some print point help
understand now paste method

class Solution(object):
    def maximalRectangle(self, matrix):
        """
        :type matrix: List[List[str]]
        :rtype: int
        """
        if not matrix or not matrix[0]: return 0
        M, N = len(matrix), len(matrix[0])
        height = [0] * N
        res = 0
        for row in matrix:
            for i in range(N):
                if row[i] == '0':
                    height[i] = 0
                else:
                    height[i] += 1
                print("height1")
                print(height)
            res = max(res, self.maxRectangleArea(height))
        return res

    def maxRectangleArea(self, height):
        if not height: return 0
        print("height")
        print(height)
        res = 0
        stack = list()
        print("stack0")
        print(stack)
        height.append(0)
        print("height2")
        print(height)
        for i in range(len(height)):
            cur = height[i]
            while stack and cur < height[stack[-1]]:
                print("stack")
                print(stack)
                w = height[stack.pop()]
                h = i if not stack else i - stack[-1] - 1
                res = max(res, w * h)
            stack.append(i)
        return res

solu = Solution()
matirx=[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
print(solu.maximalRectangle(matirx))

 

转载于:https://www.cnblogs.com/Marigolci/p/11337427.html