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.

public static int MaximalRectangle(List<List<char>> matrix)
{
int n = matrix.Count;
int m = matrix[0].Count;
int curr_area = 0;
int max_area = 0;

for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if(matrix[i][j] == '1')
{
int lastj = m;
for (int k = i; k < n && matrix[k][j] == '1'; k++)
{
for (int l = j; l < lastj; l++)
{
if (matrix[k][l] == '1')
{
curr_area = (l - j + 1) * (k - i + 1);
max_area = Math.Max(max_area, curr_area);
}
else
lastj = l;
}
}
}
}
}

return max_area;
}

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;

　　...

public static int MaximalRectangleOpt(List<List<char>> matrix)
{
int n = matrix.Count;
int m = matrix[0].Count;
int[,] dp = new int[n, m];
int curr_area = 0;
int max_area = 0;

for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (matrix[i][j] == '1')
dp[i, j] = j > 0 ? dp[i, j - 1] + 1 : 1;
else
dp[i, j] = 0;
}
}

for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
int k = i;
int min = dp[i, j];
while (k >= 0 && dp[k,j] > 0)
{
min = Math.Min(min, dp[k, j]);
curr_area = min * (i - k + 1);
max_area = Math.Max(max_area, curr_area);
k--;
}
}
}

return max_area;
}

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