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                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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