题目地址: https://leetcode.com/problems/maximal-rectangle/description/

题目描述:

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

题目大意

给出了一个二维的数组,求在这里面能够成的最大的矩形面积是多少。

解题方法

本以为会和221. Maximal Squareopen in new window做法一样使用DP解决,可是感觉状态转移方程太复杂,所以还是参考了84. Largest Rectangle in Histogramopen in new window升级版的解法。

如图所示,如果把每一行的1和它上面的1连在一起,那么就可以看成一个个站里的矩形方块。那么我们的最终目的就是找出最大面积的矩形方块,所以就是第84题的做法了,使用单调栈。

 

需要注意的是,我们使用一个height数组,保存到某一层的第i个位置为止,能向上构成的矩形的高度。而且需要对每层都做一个寻找面积的操作,最终选择所有层中能够成矩形面积最大值。

时间复杂度是O(M(N+M)),空间复杂度是O(N)。

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

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