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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