leetcode-85-最大矩形

题目描述：

方法一：动态规划+使用柱状图的优化暴力方法 O(N*2M） O(NM) N为行数

class Solution:
    def maximalRectangle(self, matrix: List[List[str]]) -> int:
        maxarea = 0

        dp = [[0] * len(matrix[0]) for _ in range(len(matrix))]
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if matrix[i][j] == '0': continue

                # compute the maximum width and update dp with it
                width = dp[i][j] = dp[i][j-1] + 1 if j else 1

                # compute the maximum area rectangle with a lower right corner at [i, j]
                for k in range(i, -1, -1):
                    width = min(width, dp[k][j])
                    maxarea = max(maxarea, width * (i-k+1))
        return maxarea

方法二：栈 参考84题 O(NM) O(M)

class Solution:
    def maximalRectangle(self, matrix: List[List[str]]) -> int:
        if not matrix: return 0
        maxarea = 0
        dp = [0 for _ in range(len(matrix[0]))]
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                dp[j] = dp[j] + 1 if matrix[i][j] == "1" else 0
            maxarea = max(maxarea,self.largestRectangleArea(dp))
        return maxarea

    def largestRectangleArea(self, heights: List[int]) -> int:
        stack = [0]
        heights = [0] + heights + [0]
        res = 0
        for i in range(len(heights)):
            while heights[stack[-1]] > heights[i]:
                tmp = stack.pop()
                res = max(res, (i - stack[-1] - 1) * heights[tmp])
            stack.append(i)
        return res

方法三：动态规划  O(NM)

class Solution:
    def maximalRectangle(self, matrix: List[List[str]]) -> int:
        if not matrix or not matrix[0]: return 0
        row = len(matrix)
        col = len(matrix[0])
        left_j = [-1] * col
        right_j = [col] * col
        height_j = [0] * col
        res = 0
        for i in range(row):
            cur_left = -1
            cur_right = col

            for j in range(col):
                if matrix[i][j] == "1":
                    height_j[j] += 1
                else:
                    height_j[j] = 0

            for j in range(col):
                if matrix[i][j] == "1":
                    left_j[j] = max(left_j[j], cur_left)
                else:
                    left_j[j] = -1
                    cur_left = j

            for j in range(col - 1, -1, -1):
                if matrix[i][j] == "1":
                    right_j[j] = min(right_j[j], cur_right)
                else:
                    right_j[j] = col
                    cur_right = j
            for j in range(col):
                res = max(res, (right_j[j] - left_j[j] - 1) * height_j[j])
        return res