[LeetCode] 85. Maximal Rectangle

题：https://leetcode.com/problems/maximal-rectangle/description/

题目

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

解法

DP思想，对于每个 i,j 求它的 左右边界和累加的高度然后算法矩形的面积，由于高度为可以累加的，左右边界是继承的，即，上个row的i,j 的边界和现在i,j的边界关系是上个row的边界一定包含当前row边界。
通过左右各次扫描能得出相应的边界。
可以和 [LeetCode] 79. Word Search 对比下
故从上个row中提取最好，可能成为当前row最好的，这就是题目中的状态转移。

https://leetcode.com/problems/maximal-rectangle/discuss/29054/Share-my-DP-solution

The DP solution proceeds row by row, starting from the first row. Let the maximal rectangle area at row i and column j be computed by [right(i,j) - left(i,j)]*height(i,j).

All the 3 variables left, right, and height can be determined by the information from previous row, and also information from the current row. So it can be regarded as a DP solution. The transition equations are:

left(i,j) = max(left(i-1,j), cur_left), cur_left can be determined from the current row

right(i,j) = min(right(i-1,j), cur_right), cur_right can be determined from the current row

height(i,j) = height(i-1,j) + 1, if matrix[i][j]==’1’;

height(i,j) = 0, if matrix[i][j]==’0’

The code is as below. The loops can be combined for speed but I separate them for more clarity of the algorithm.

code

class Solution:
    def maximalRectangle(self, matrix):
        """
        :type matrix: List[List[str]]
        :rtype: int
        """
        m = len(matrix)
        if m == 0:
            return 0
        n = len(matrix[0])

        bleft = {}
        bright ={}
        heights = {}
        for i in range(n):
            bleft[i] = -1
            bright[i] = n
            heights[i] = 0

        maxm = 0
        for i in range(m):
            cur_left = -1
            cur_right = n
            for j in range(n):
                if matrix[i][j] == "1":
                    bleft[j] = max(bleft[j],cur_left)
                    heights[j] += 1
                else:
                    bleft[j] = -1
                    cur_left = j
                    heights[j] = 0

            for j in range(n-1,0-1,-1):
                if matrix[i][j] == "1":
                    bright[j] = min(bright[j],cur_right)
                else:
                    bright[j] = n
                    cur_right = j

            for j in range(n):
                maxm = max(maxm,heights[j]*(bright[j]-bleft[j]-1))

        return maxm