【leetcode刷题笔记】Maximal Rectangle

transfer[0][j] = matrix[0][j] - ‘0‘;

transfer[i][j] = (matrix[i][j] == ‘0‘ ? 0: transfer[i-1][j] + 1); (i >= 1)

 1 public class Solution {
 2     public int largestRectangleArea(int[] height) {
 3         if(height == null || height.length == 0)
 4             return 0;
 5         Stack<Integer> index = new Stack<Integer>();
 6         int totalMax = 0;
 7         ArrayList<Integer> newHeight = new ArrayList<Integer>();
 8         for(int i:height) newHeight.add(i);
 9         newHeight.add(0);
10         
11         
12         for(int i = 0;i < newHeight.size();i++){
13             if(index.isEmpty() || newHeight.get(i) >= newHeight.get(index.peek()))
14                 index.push(i);
15             else{
16                 int top = index.pop();
17                 totalMax = Math.max(totalMax,newHeight.get(top) * (index.isEmpty()?i:i-index.peek()-1));
18                 i--;
19             }
20         }
21         
22         return totalMax;
23     }
24     public int maximalRectangle(char[][] matrix) {
25         if(matrix == null || matrix.length == 0)
26             return 0;
27         int m = matrix.length;
28         int n = matrix[0].length;
29         int totalMax = 0;
30         int[][] transfer = new int[m][n];
31         
32         //transform matrix to histogram in a new matrix
33         for(int i = 0;i < n;i++)
34             transfer[0][i] = matrix[0][i]- ‘0‘;
35         for(int i = 1;i < m;i++){
36             for(int j = 0;j < n;j++){
37                 if(matrix[i][j] == ‘0‘ )
38                     transfer[i][j] = 0;
39                 else {
40                     transfer[i][j] = transfer[i-1][j] + 1; 
41                 }
42             }
43         }
44         
45         //Using histogram to find the biggest rectangle
46         for(int i = 0;i < m;i++){
47             totalMax = Math.max(totalMax, largestRectangleArea(transfer[i]));
48         }
49         
50         return totalMax;
51     }
52 }