• 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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