【leetcode刷题笔记】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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题解，很巧妙的一道题，对于一个0-1矩阵，它的每一行及以上都可以看作一个直方图（如下图所示），利用Largest Rectangle in Histogram的方法，可以在O(n)的时间搜索出这一行及以上的直方图中面积最大的矩形，对矩阵的每一行依次做这个操作，就可以在O(n²)的时间搜索出最大的矩形了。

将原矩阵转换成直方图矩阵的方法：设用transfer矩阵存放转换后的直方图，则

transfer[0][j] = matrix[0][j] - '0';



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

代码如下：

 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 }