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[leetcode]85. Maximal Rectangle 最大矩形

时间:2019-05-31 16:33:58      阅读:91      评论:0      收藏:0      [点我收藏+]

标签:return   https   follow   class   ogr   else   pop   rect   ++   

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

Example:

Input:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
Output: 6

 

题意

给定一个01矩阵,找出其中全由1构成的最大矩形。

 

思路

将所有1连接的部分看成一个bar height,这个题就可以看成是[leetcode]84. Largest Rectangle in Histogram直方图中的最大矩形

的followup,进行代码复用

 

代码

 1 class Solution {
 2     public int maximalRectangle(char[][] matrix) {
 3        if(matrix.length==0 || matrix[0].length==0) return 0;
 4        int row = matrix.length;
 5         int col = matrix[0].length;
 6         int [] heights = new int[col];
 7         int area = 0;
 8         for(int i = 0; i< row; i++){
 9             for(int j = 0; j< col; j++){
10                 if(matrix[i][j]==‘1‘){
11                     heights[j]++;
12                 }else{
13                     heights[j]=0;
14                 }
15                 
16             }     
17            area = Math.max(area,  largestRectangle(heights));
18         }
19          return area;     
20     }
21     
22     public int largestRectangle(int[] heights ){
23         int area = 0;
24         Stack<Integer> s = new Stack<>();
25         for(int i = 0; i<=heights.length;){
26             int value = i<heights.length ? heights[i] : 0;
27             if(s.isEmpty() || value > heights[s.peek()]){
28                 s.push(i);
29                 i++;
30             }else{
31                 int temp = s.pop();
32                 area = Math.max(area, heights[temp]*(s.isEmpty()? i: i-s.peek()-1));
33             }// end else 
34         }// end for 
35         return area; 
36     }
37 }

 

[leetcode]85. Maximal Rectangle 最大矩形

标签:return   https   follow   class   ogr   else   pop   rect   ++   

原文地址:https://www.cnblogs.com/liuliu5151/p/10955734.html

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