LeetCode Online Judge 题目C# 练习 - Maximal Rectangle

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

 1         public static int MaximalRectangle(List<List<char>> matrix)

 2         {

 3             int n = matrix.Count;

 4             int m = matrix[0].Count;

 5             int curr_area = 0;

 6             int max_area = 0;

 7 

 8             for (int i = 0; i < n; i++)

 9             {

10                 for (int j = 0; j < m; j++)

11                 {

12                     if(matrix[i][j] == '1')

13                     {

14                         int lastj = m;

15                         for (int k = i; k < n && matrix[k][j] == '1'; k++)

16                         {

17                             for (int l = j; l < lastj; l++)

18                             {

19                                 if (matrix[k][l] == '1')

20                                 {

21                                     curr_area = (l - j + 1) * (k - i + 1);

22                                     max_area = Math.Max(max_area, curr_area);

23                                 }

24                                 else

25                                     lastj = l;

26                             }

27                         }

28                     }

29                 }

30             }

31 

32             return max_area;

33         }

代码分析：

　　Brute Force, O(n⁴). 每一个是1的格子，都从上至下，从左往右，去看最大面积。lastj变量是存放最小的列。

　　例如：　　

1	1	0	1
1	0	1	1
1	1	1	1
0	0	1	1

　　[0,0] :  第一行: curr_area: 2, max_area: 2, lastj = 2

　　　　　　第二行: curr_area: 2, max_area: 2, lastj = 1

　　　　　　第三行: curr_area: 3, max_area: 3, lastj = 1

　　　　　　max_area = 3;

　　...

 

 1         public static int MaximalRectangleOpt(List<List<char>> matrix)

 2         {

 3             int n = matrix.Count;

 4             int m = matrix[0].Count;

 5             int[,] dp = new int[n, m];

 6             int curr_area = 0;

 7             int max_area = 0;

 8             

 9             for (int i = 0; i < n; i++)

10             {

11                 for (int j = 0; j < m; j++)

12                 {

13                     if (matrix[i][j] == '1')

14                         dp[i, j] = j > 0 ? dp[i, j - 1] + 1 : 1;

15                     else

16                         dp[i, j] = 0;

17                 }

18             }

19 

20             for (int i = 0; i < n; i++)

21             {

22                 for (int j = 0; j < m; j++)

23                 {

24                     int k = i;

25                     int min = dp[i, j];

26                     while (k >= 0 && dp[k,j] > 0)

27                     {

28                         min = Math.Min(min, dp[k, j]);

29                         curr_area = min * (i - k + 1);

30                         max_area = Math.Max(max_area, curr_area);

31                         k--;

32                     }

33                 }

34             }

35 

36             return max_area;

37         }

代码分析：

　　利用DP，第一次n²循环,先预处理。然后再O(n³)计算最大面积。

　　例如：

1	1	0	1
1	0	1	1
1	1	1	1
0	0	1	1

　　DP处理后生成新的2D- array

1	2	0	1
1	0	1	2
1	2	3	4
0	0	1	2

 

　　每个格子相应的最大面积

1	2	0	1
2	0	1	2
3	2	3	4
0	0	3	6