leetcode: 85. Maximal Rectangle

# Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
#
# For example, given the following matrix:
#
# 1 0 1 0 0
# 1 0 1 1 1
# 1 1 1 1 1
# 1 0 0 1 0
# Return 6.

如果我们把每一行看成x坐标，那高度就是 从那一行开始往上数的1的个数 。

带入我们的maxAreaInHist方法，在O(n2)时间内就可以求出每一行形成的“柱状图”的最大矩形面积了。
它们之中最大的，就是我们要的答案。

class Solution():
    def maximalRectangle(self, x):
        maxArea = 0
        if not x:
            return 0
        vector = [0]* len(x[0])
        for r in x:
            for i, c in enumerate(r):
                if c == '0':
                    vector[i]=0
                else:
                    vector[i]+=1
            maxArea = max(maxArea, self.largestRectangleArea(vector[:]))
        return maxArea

    def largestRectangleArea(self, xs):
        xs.insert(0,0)
        xs.append(0)
        stack, maxArea = [], 0
        for i, x in enumerate(xs):
            if not stack or xs[stack[-1]]<=x:
                stack.append(i)
            else:
                while stack and xs[stack[-1]]>x:
                    maxArea = max(maxArea, xs[stack[-1]]*(i-stack[-2]-1))
                    stack.pop()
                stack.append(i)
        return maxArea


if __name__ == "__main__":
    matrix = ["01101", "11010", "01110", "11110", "11111", "00000"]
    assert Solution().maximalRectangle(matrix) == 9