[Leetcode] 85. Maximal Rectangle rectangle maximum [leetcode] 84. Largest Rectangle in Histogram histogram maximum rectangular

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

 

The meaning of problems

Given a 01 matrix to find out where the whole by one of the largest rectangles.

 

Thinking

All as part of a connecting bar height, this problem can be seen as [leetcode] 84. Largest Rectangle in Histogram histogram maximum rectangular

The followup, code reuse

 

Code

 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 }