leetcode85. 最大矩形

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兔老大 发表于 2021/04/23 00:56:35 2021/04/23
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【摘要】 给定一个仅包含 0 和 1 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。 示例: 输入: [   ["1","0","1","0","0"],   ["1","0","1","1","1"],   ["1","1","1","1","1"],   ["1","0","0","1","0"] ] 输出: 6 ...

给定一个仅包含 0 和 1 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。

示例:

输入:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
输出: 6

和leetcode84没什么区别,就是一行一行更新出“高度”,一行一行的跑一遍单调栈即可。


      class Solution {
      public int leetcode84(int[] heights) {
       Stack < Integer > stack = new Stack < > ();
       stack.push(-1);
      int maxarea = 0;
      for (int i = 0; i < heights.length; ++i) {
      while (stack.peek() != -1 && heights[stack.peek()] >= heights[i])
       maxarea = Math.max(maxarea, heights[stack.pop()] * (i - stack.peek() - 1));
       stack.push(i);
       }
      while (stack.peek() != -1)
       maxarea = Math.max(maxarea, heights[stack.pop()] * (heights.length - stack.peek() -1));
      return maxarea;
       }
      public int maximalRectangle(char[][] matrix) {
      if (matrix.length == 0) return 0;
      int maxarea = 0;
      int[] dp = new int[matrix[0].length];
      for(int i = 0; i < matrix.length; i++) {
      for(int j = 0; j < matrix[0].length; j++) {
       dp[j] = matrix[i][j] == '1' ? dp[j] + 1 : 0;
       }
       maxarea = Math.max(maxarea, leetcode84(dp));
       } return maxarea;
       }
      }
  
 

文章来源: fantianzuo.blog.csdn.net,作者:兔老大RabbitMQ,版权归原作者所有,如需转载,请联系作者。

原文链接:fantianzuo.blog.csdn.net/article/details/104027791

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