我正在试图弄清楚如何为Kadane的2D Matrix算法实现C＃代码。我在这里找到了一维版本：
但我想要一个2D版本。基本上，给定一个正数和负数的矩阵N x N，我需要找到一个子矩阵，其中所有元素的总和将是最大的。
答案 0 :(得分：3)
想出来。对于那些有兴趣的人
static void Main(string[] args)
{
int[,] array2D = new int[,]
{
{ 1, 2 },
{ -3, 4 },
{ 5, -6 },
{ -7, -8 }
};
var max = GetMaxMatrix(array2D);
Console.WriteLine(max);
}
public static int GetMaxMatrix(int[,] original)
{
int maxArea = int.MinValue;
int rowCount = original.GetLength(0);
int columnCount = original.GetLength(1);
int[,] matrix = PrecomputeMatrix(original);
for (int row1 = 0; row1 < rowCount; row1++)
{
for (int row2 = row1; row2 < rowCount; row2++)
{
for (int col1 = 0; col1 < columnCount; col1++)
{
for (int col2 = col1; col2 < columnCount; col2++)
{
maxArea = Math.Max(maxArea, ComputeSum(matrix, row1, row2, col1, col2));
}
}
}
}
return maxArea;
}
private static int[,] PrecomputeMatrix(int[,] matrix)
{
var sumMatrix = new int[matrix.GetLength(0), matrix.GetLength(1)];
for (int i = 0; i < matrix.GetLength(0); i++)
{
for (int j = 0; j < matrix.GetLength(1); j++)
{
if (i == 0 && j == 0)
{ // first cell
sumMatrix[i, j] = matrix[i, j];
}
else if (j == 0)
{ // cell in first column
sumMatrix[i, j] = sumMatrix[i - 1, j] + matrix[i, j];
}
else if (i == 0)
{ // cell in first row
sumMatrix[i, j] = sumMatrix[i, j - 1] + matrix[i, j];
}
else
{
sumMatrix[i, j] = sumMatrix[i - 1, j] +
sumMatrix[i, j - 1] - sumMatrix[i - 1, j - 1] + matrix[i, j];
}
}
}
return sumMatrix;
}
private static int ComputeSum(int[,] sumMatrix, int i1, int i2, int j1, int j2)
{
if (i1 == 0 && j1 == 0)
{ // starts at row 0, column 0
return sumMatrix[i2, j2];
}
else if (i1 == 0)
{ // start at row 0
return sumMatrix[i2, j2] - sumMatrix[i2, j1 - 1];
}
else if (j1 == 0)
{ // start at column 0
return sumMatrix[i2, j2] - sumMatrix[i1 - 1, j2];
}
else
{
return sumMatrix[i2, j2] - sumMatrix[i2, j1 - 1]
- sumMatrix[i1 - 1, j2] + sumMatrix[i1 - 1, j1 - 1];
}
}
答案 1 :(得分：3)
这是一个太诱人的问题，没有触及： - ）
我找到了问题here的C / C ++解决方案，基本上我已将其翻译成C＃。解决方案适用于int：s，但将其转换为您喜欢的数字类型应该相当容易。您可能还希望以某种方式返回findMaxSum的结果，我将其留作练习。
// Driver program to test 2D Kadane method
void Main()
{
int[,] M = {{ 1, 2, -1, -4, -20},
{-8, -3, 4, 2, 1},
{ 3, 8, 10, 1, 3},
{-4, -1, 1, 7, -6}};
findMaxSum(M);
}
// Implementation of Kadane's algorithm for 1D array. The function returns the
// maximum sum and stores starting and ending indexes of the maximum sum subarray
// at addresses pointed by start and finish pointers respectively.
int kadane(int[] arr, out int start, out int finish, int n)
{
// initialize sum, maxSum
int sum = 0;
int maxSum = Int32.MinValue;
// Just some initial value to check for all negative values case
start = -1;
finish = -1;
int local_start = 0;
for (int i = 0; i < n; ++i)
{
sum += arr[i];
if (sum < 0)
{
sum = 0;
local_start = i+1;
}
else if (sum > maxSum)
{
maxSum = sum;
start = local_start;
finish = i;
}
}
// There is at-least one non-negative number
if (finish != -1)
return maxSum;
// Special Case: When all numbers in arr[] are negative
maxSum = arr[0];
start = finish = 0;
// Find the maximum element in array
for (int i = 1; i < n; i++)
{
if (arr[i] > maxSum)
{
maxSum = arr[i];
start = finish = i;
}
}
return maxSum;
}
// The main function that finds maximum sum rectangle in M[][]
void findMaxSum(int[,] M)
{
int ROW = M.GetLength(0);
int COL = M.GetLength(1);
// Variables to store the final output
int maxSum = Int32.MinValue;
int finalLeft = -1, finalRight = -1, finalTop = -1, finalBottom = -1;
// Set the left column
for (int left = 0; left < COL; ++left)
{
// Initialize all elements of temp as 0
int start, finish;
int[] temp = new int[ROW];
// Set the right column for the left column set by outer loop
for (int right = left; right < COL; ++right)
{
// Calculate sum between current left and right for every row 'i'
for (int i = 0; i < ROW; ++i)
temp[i] += M[i, right];
// Find the maximum sum subarray in temp[]. The kadane() function
// also sets values of start and finish. So 'sum' is sum of
// rectangle between (start, left) and (finish, right) which is the
// maximum sum with boundary columns strictly as left and right.
int sum = kadane(temp, out start, out finish, ROW);
// Compare sum with maximum sum so far. If sum is more, then update
// maxSum and other output values
if (sum > maxSum)
{
maxSum = sum;
finalLeft = left;
finalRight = right;
finalTop = start;
finalBottom = finish;
}
}
}
// Print final values
Console.WriteLine("(Top, Left) ({0},{1})", finalTop, finalLeft);
Console.WriteLine("(Bottom, Right) ({0},{1})", finalBottom, finalRight);
Console.WriteLine("Max sum is: {0}\n", maxSum);
}