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
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