LeetCode: Maximal Rectangle 解题报告

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

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SOLUTION 1:

 1 public class Solution {
 2     public int maximalRectangle(char[][] matrix) {
 3         if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
 4             return 0;
 5         }
 6         
 7         int rows = matrix.length;
 8         int cols = matrix[0].length;
 9         
10         int[][] h = new int[rows][cols];
11         
12         int max = 0;
13         
14         for (int i = 0; i < rows; i++) {
15             for (int j = 0; j < cols; j++) {
16                 
17                 h[i][j] = matrix[i][j] == '1' ? 1: 0;
18                 
19                 if (i != 0 && h[i][j] != 0) {
20                     h[i][j] = h[i - 1][j] + 1;
21                 }
22                 
23                 if (j == cols - 1) {
24                     max = Math.max(max, maxArea(h[i]));
25                 }
26             }
27         }
28         
29         return max;
30     }
31     
32     public int maxArea(int[] h) {
33         Stack<Integer> s = new Stack<Integer>();
34         
35         int max = 0;
36         int i = 0;
37         
38         // 注意，这里使用<= 因为当i到达最后，需要计算一次。
39         while (i <= h.length) {
40             // 
41             if (s.isEmpty() || i < h.length && h[i] >= h[s.peek()]) {
42                 s.push(i);
43                 i++;
44             } else {
45                 int height = h[s.pop()];
46                 int width = s.isEmpty() ? i: i - s.peek() - 1;
47                 max = Math.max(max, height * width);
48             }
49         }
50         
51         return max;
52     }
53 }

 2015.1.3 redo:

 1 public class Solution {
 2     public int maximalRectangle(char[][] matrix) {
 3         if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
 4             return 0;
 5         }
 6         
 7         int rows = matrix.length;
 8         int cols = matrix[0].length;
 9         int[][] h = new int[matrix.length][matrix[0].length];
10         
11         int maxArea = 0;
12         
13         for (int i = 0; i < rows; i++) {
14             for (int j = 0; j < cols; j++) {
15                 if (matrix[i][j] == '0') {
16                     h[i][j] = 0;
17                 } else {
18                     if (i == 0) {
19                         h[i][j] = 1;    
20                     } else {
21                         h[i][j] = 1 + h[i - 1][j];
22                     }
23                 }
24                 
25                 if (j == cols - 1) {
26                     // the last element of every row.
27                     int maxRec = maxAreaInRow(h[i]);
28                     maxArea = Math.max(maxArea, maxRec);
29                 }
30             }
31         }
32         
33         return maxArea;
34     }
35     
36     public int maxAreaInRow(int[] h) {
37         Stack<Integer> s = new Stack<Integer>();
38         
39         int i = 0;
40         int max = 0;
41         while (i <= h.length) {
42             // the current height is higher or equal than the one in the stack.
43             if (s.isEmpty() || i < h.length && h[i] >= h[s.peek()]) {
44                 s.push(i);
45                 i++;
46             } else {
47                 // get the area.
48                 int hight = h[s.pop()];
49                 int w = s.isEmpty() ? i: i - s.peek() - 1;
50                 max = Math.max(max, hight * w);
51             }
52         }
53         
54         return max;
55     }
56 }

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