【每日一题-LC85】最大矩形

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

Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 6
Explanation: The maximal rectangle is shown in the above picture.

[Python3 code]

先每列求和，然后对每一行做分析，取连续都不为0的列即为矩形的长度，高度为这几列的最小值，记录面积，最后返回最大面积。

其实就是每行做“最大储水池”的LC

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)
        ans = 0
        for row in matrix:
            for i in range(n):
                height[i] = height[i] + 1 if row[i] == '1' else 0
            stack = [-1]
            for i in range(n + 1):
                while height[i] < height[stack[-1]]:
                    h = height[stack.pop()]
                    w = i - 1 - stack[-1]
                    ans = max(ans, h * w)
                stack.append(i)
        return ans

class Solution:
    def maximalRectangle(self, matrix):
        if not matrix:
            return 0
        h, w = len(matrix), len(matrix[0])
        m = [[0]*w for _ in range(h)]
        for j in range(h):
            for i in range(w):
                if matrix[j][i] == '1':
                    m[j][i] = m[j-1][i] + 1
        return max(self.largestRectangleArea(row) for row in m)

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

时间复杂度分析

O(n^2)