【Leetcode】Maximal Rectangle

最新推荐文章于 2025-08-08 21:26:37 发布

转载最新推荐文章于 2025-08-08 21:26:37 发布 · 193 阅读

文章标签：

#leetcode #Java #LinkedList #String

本文介绍了一种求解二维二进制矩阵中全为1的最大矩形面积的算法。通过迭代更新高度数组并利用直方图最大矩形面积的方法，实现了高效求解。文章提供了完整的Java代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.

Java:

1. http://blog.youkuaiyun.com/linhuanmars/article/details/24444491 根据原来Largest Rectangle in Histogram的改过来

2. http://blog.youkuaiyun.com/muscler/article/details/23641887

public class Solution {
    public int maximalRectangle(char[][] matrix) {
          if(matrix==null||matrix.length==0||matrix[0].length==0) return 0;
          int maxArea=0;
          int[] height=new int[matrix[0].length];
          for(int i=0;i<matrix.length;i++)
          {
              for(int j=0;j<matrix[0].length;j++)
              {
                  height[j]=matrix[i][j]=='0'?0:height[j]+1;
              }
              maxArea=Math.max(maxArea,largestRectangleArea(height));
          }
          return maxArea;
    }  
    public int largestRectangleArea(int[] height) {  
    if(height==null || height.length==0)  
    {  
        return 0;  
    }  
    int maxArea = 0;  
    LinkedList<Integer> stack = new LinkedList<Integer>();  
    for(int i=0;i<height.length;i++)  
    {  
          
        while(!stack.isEmpty() && height[i]<=height[stack.peek()])  
        {  
            int index = stack.pop();  
            int curArea = stack.isEmpty()?i*height[index]:(i-stack.peek()-1)*height[index];  
            maxArea = Math.max(maxArea,curArea);  
        }  
        stack.push(i);  
    }  
    while(!stack.isEmpty())  
    {  
        int index = stack.pop();  
        int curArea = stack.isEmpty()?height.length*height[index]:(height.length-stack.peek()-1)*height[index];  
        maxArea = Math.max(maxArea,curArea);              
        
    }  
    return maxArea;  
        
    }  
}