先决条件： Kadane 算法
给定一个维度为N*M的二维数组arr[][] ，任务是从矩阵arr[][] 中找到最大和子矩阵。
例子：
Input: arr[][] = {{0, -2, -7, 0 }, { 9, 2, -6, 2 }, { -4, 1, -4, 1 }, { -1, 8, 0, -2}}
Output: 15
Explanation: The submatrix {{9, 2}, {-4, 1}, {-1, 8}} has a sum 15, which is the maximum sum possible.
Input: arr[][] = {{1, 2}, {-5, -7}}
Output: 3
朴素方法：最简单的方法是从给定矩阵生成所有可能的子矩阵并计算它们的总和。最后，打印获得的最大和。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find maximum sum submatrix
void maxSubmatrixSum(
vector > matrix)
{
// Stores the number of rows
// and columns in the matrix
int r = matrix.size();
int c = matrix[0].size();
// Stores maximum submatrix sum
int maxSubmatrix = 0;
// Take each row as starting row
for (int i = 0; i < r; i++) {
// Take each column as the
// starting column
for (int j = 0; j < c; j++) {
// Take each row as the
// ending row
for (int k = i; k < r; k++) {
// Take each column as
// the ending column
for (int l = j; l < c; l++) {
// Stores the sum of submatrix
// having topleft index(i, j)
// and bottom right index (k, l)
int sumSubmatrix = 0;
// Iterate the submatrix
// row-wise and calculate its sum
for (int m = i; m <= k; m++) {
for (int n = j; n <= l; n++) {
sumSubmatrix += matrix[m][n];
}
}
// Update the maximum sum
maxSubmatrix
= max(maxSubmatrix,
sumSubmatrix);
}
}
}
}
// Print the answer
cout << maxSubmatrix;
}
// Driver Code
int main()
{
vector > matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
maxSubmatrixSum(matrix);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find maximum sum submatrix
static void maxSubmatrixSum(int[][] matrix)
{
// Stores the number of rows
// and columns in the matrix
int r = matrix.length;
int c = matrix[0].length;
// Stores maximum submatrix sum
int maxSubmatrix = 0;
// Take each row as starting row
for (int i = 0; i < r; i++) {
// Take each column as the
// starting column
for (int j = 0; j < c; j++) {
// Take each row as the
// ending row
for (int k = i; k < r; k++) {
// Take each column as
// the ending column
for (int l = j; l < c; l++) {
// Stores the sum of submatrix
// having topleft index(i, j)
// and bottom right index (k, l)
int sumSubmatrix = 0;
// Iterate the submatrix
// row-wise and calculate its sum
for (int m = i; m <= k; m++) {
for (int n = j; n <= l; n++) {
sumSubmatrix += matrix[m][n];
}
}
// Update the maximum sum
maxSubmatrix
= Math.max(maxSubmatrix,
sumSubmatrix);
}
}
}
}
// Print the answer
System.out.println(maxSubmatrix);
}
// Driver Code
public static void main(String[] args)
{
int[][] matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
maxSubmatrixSum(matrix);
}
}
// This code is contributed by susmitakundugoaldanga.Python3
# Python3 program for the above approach
# Function to find maximum sum submatrix
def maxSubmatrixSum(matrix):
# Stores the number of rows
# and columns in the matrix
r = len(matrix)
c = len(matrix[0])
# Stores maximum submatrix sum
maxSubmatrix = 0
# Take each row as starting row
for i in range(r):
# Take each column as the
# starting column
for j in range(c):
# Take each row as the
# ending row
for k in range(i, r):
# Take each column as
# the ending column
for l in range(j, c):
# Stores the sum of submatrix
# having topleft index(i, j)
# and bottom right index (k, l)
sumSubmatrix = 0
# Iterate the submatrix
# row-wise and calculate its sum
for m in range(i, k + 1):
for n in range(j, l + 1):
sumSubmatrix += matrix[m][n]
# Update the maximum sum
maxSubmatrix= max(maxSubmatrix, sumSubmatrix)
# Prthe answer
print (maxSubmatrix)
# Driver Code
if __name__ == '__main__':
matrix = [ [ 0, -2, -7, 0 ],
[ 9, 2, -6, 2 ],
[ -4, 1, -4, 1 ],
[ -1, 8, 0, -2 ] ]
maxSubmatrixSum(matrix)
# This code is contributed by mohit kumar 29.C#
// C# program to implement
// the above approach
using System;
public class GFG
{
// Function to find maximum sum submatrix
static void maxSubmatrixSum(int[,] matrix)
{
// Stores the number of rows
// and columns in the matrix
int r = matrix.GetLength(0);
int c = matrix.GetLength(1);
// Stores maximum submatrix sum
int maxSubmatrix = 0;
// Take each row as starting row
for (int i = 0; i < r; i++) {
// Take each column as the
// starting column
for (int j = 0; j < c; j++) {
// Take each row as the
// ending row
for (int k = i; k < r; k++) {
// Take each column as
// the ending column
for (int l = j; l < c; l++) {
// Stores the sum of submatrix
// having topleft index(i, j)
// and bottom right index (k, l)
int sumSubmatrix = 0;
// Iterate the submatrix
// row-wise and calculate its sum
for (int m = i; m <= k; m++) {
for (int n = j; n <= l; n++) {
sumSubmatrix += matrix[m, n];
}
}
// Update the maximum sum
maxSubmatrix
= Math.Max(maxSubmatrix,
sumSubmatrix);
}
}
}
}
// Print the answer
Console.WriteLine(maxSubmatrix);
}
// Driver Code
public static void Main(String []args)
{
int[,] matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
maxSubmatrixSum(matrix);
}
}
// This code is contributed by sanjoy_62.Javascript
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find maximum continuous
// maximum sum in the array
int kadane(vector v)
{
// Stores current and maximum sum
int currSum = 0;
int maxSum = INT_MIN;
// Traverse the array v
for (int i = 0;
i < (int)v.size(); i++) {
// Add the value of the
// current element
currSum += v[i];
// Update the maximum sum
if (currSum > maxSum) {
maxSum = currSum;
}
if (currSum < 0) {
currSum = 0;
}
}
// Return the maximum sum
return maxSum;
}
// Function to find the maximum
// submatrix sum
void maxSubmatrixSum(
vector > A)
{
// Store the rows and columns
// of the matrix
int r = A.size();
int c = A[0].size();
// Create an auxiliary matrix
int** prefix = new int*[r];
// Traverse the matrix, prefix
// and initialize it will all 0s
for (int i = 0; i < r; i++) {
prefix[i] = new int;
for (int j = 0; j < c; j++) {
prefix[i][j] = 0;
}
}
// Calculate prefix sum of all
// rows of matrix A[][] and
// store in matrix prefix[]
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
// Update the prefix[][]
if (j == 0)
prefix[i][j] = A[i][j];
else
prefix[i][j] = A[i][j]
+ prefix[i][j - 1];
}
}
// Store the maximum submatrix sum
int maxSum = INT_MIN;
// Iterate for starting column
for (int i = 0; i < c; i++) {
// Iterate for last column
for (int j = i; j < c; j++) {
// To store current array
// elements
vector v;
// Traverse every row
for (int k = 0; k < r; k++) {
// Store the sum of the
// kth row
int el = 0;
// Update the prefix
// sum
if (i == 0)
el = prefix[k][j];
else
el = prefix[k][j]
- prefix[k][i - 1];
// Push it in a vector
v.push_back(el);
}
// Update the maximum
// overall sum
maxSum = max(maxSum, kadane(v));
}
}
// Print the answer
cout << maxSum << "\n";
}
// Driver Code
int main()
{
vector > matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
// Function Call
maxSubmatrixSum(matrix);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find maximum continuous
// maximum sum in the array
static int kadane(Vector v)
{
// Stores current and maximum sum
int currSum = 0;
int maxSum = Integer.MIN_VALUE;
// Traverse the array v
for (int i = 0;
i < (int)v.size(); i++)
{
// Add the value of the
// current element
currSum += v.get(i);
// Update the maximum sum
if (currSum > maxSum)
{
maxSum = currSum;
}
if (currSum < 0)
{
currSum = 0;
}
}
// Return the maximum sum
return maxSum;
}
// Function to find the maximum
// submatrix sum
static void maxSubmatrixSum(int [][]A)
{
// Store the rows and columns
// of the matrix
int r = A.length;
int c = A[0].length;
// Create an auxiliary matrix
int [][]prefix = new int[r][];
// Traverse the matrix, prefix
// and initialize it will all 0s
for (int i = 0; i < r; i++) {
prefix[i] = new int;
for (int j = 0; j < c; j++) {
prefix[i][j] = 0;
}
}
// Calculate prefix sum of all
// rows of matrix A[][] and
// store in matrix prefix[]
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
// Update the prefix[][]
if (j == 0)
prefix[i][j] = A[i][j];
else
prefix[i][j] = A[i][j]
+ prefix[i][j - 1];
}
}
// Store the maximum submatrix sum
int maxSum = Integer.MIN_VALUE;
// Iterate for starting column
for (int i = 0; i < c; i++) {
// Iterate for last column
for (int j = i; j < c; j++) {
// To store current array
// elements
Vector v = new Vector();
// Traverse every row
for (int k = 0; k < r; k++) {
// Store the sum of the
// kth row
int el = 0;
// Update the prefix
// sum
if (i == 0)
el = prefix[k][j];
else
el = prefix[k][j]
- prefix[k][i - 1];
// Push it in a vector
v.add(el);
}
// Update the maximum
// overall sum
maxSum = Math.max(maxSum, kadane(v));
}
}
// Print the answer
System.out.print(maxSum+ "\n");
}
// Driver Code
public static void main(String[] args)
{
int [][]matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
// Function Call
maxSubmatrixSum(matrix);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program for the above approach
import sys
# Function to find maximum continuous
# maximum sum in the array
def kadane(v):
# Stores current and maximum sum
currSum = 0
maxSum = -sys.maxsize - 1
# Traverse the array v
for i in range(len(v)):
# Add the value of the
# current element
currSum += v[i]
# Update the maximum sum
if (currSum > maxSum):
maxSum = currSum
if (currSum < 0):
currSum = 0
# Return the maximum sum
return maxSum
# Function to find the maximum
# submatrix sum
def maxSubmatrixSum(A):
# Store the rows and columns
# of the matrix
r = len(A)
c = len(A[0])
# Create an auxiliary matrix
# Traverse the matrix, prefix
# and initialize it will all 0s
prefix = [[0 for i in range(c)]
for j in range(r)]
# Calculate prefix sum of all
# rows of matrix A[][] and
# store in matrix prefix[]
for i in range(r):
for j in range(c):
# Update the prefix[][]
if (j == 0):
prefix[i][j] = A[i][j]
else:
prefix[i][j] = A[i][j] + prefix[i][j - 1]
# Store the maximum submatrix sum
maxSum = -sys.maxsize - 1
# Iterate for starting column
for i in range(c):
# Iterate for last column
for j in range(i, c):
# To store current array
# elements
v = []
# Traverse every row
for k in range(r):
# Store the sum of the
# kth row
el = 0
# Update the prefix
# sum
if (i == 0):
el = prefix[k][j]
else:
el = prefix[k][j] - prefix[k][i - 1]
# Push it in a vector
v.append(el)
# Update the maximum
# overall sum
maxSum = max(maxSum, kadane(v))
# Print the answer
print(maxSum)
# Driver Code
matrix = [ [ 0, -2, -7, 0 ],
[ 9, 2, -6, 2 ],
[ -4, 1, -4, 1 ],
[ -1, 8, 0, -2 ] ]
# Function Call
maxSubmatrixSum(matrix)
# This code is contributed by rag2127C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Function to find maximum continuous
// maximum sum in the array
static int kadane(List v)
{
// Stores current and maximum sum
int currSum = 0;
int maxSum = int.MinValue;
// Traverse the array v
for (int i = 0;
i < (int)v.Count; i++)
{
// Add the value of the
// current element
currSum += v[i];
// Update the maximum sum
if (currSum > maxSum)
{
maxSum = currSum;
}
if (currSum < 0)
{
currSum = 0;
}
}
// Return the maximum sum
return maxSum;
}
// Function to find the maximum
// submatrix sum
static void maxSubmatrixSum(int [,]A)
{
// Store the rows and columns
// of the matrix
int r = A.GetLength(0);
int c = A.GetLength(1);
// Create an auxiliary matrix
int [,]prefix = new int[r,c];
// Traverse the matrix, prefix
// and initialize it will all 0s
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
prefix[i,j] = 0;
}
}
// Calculate prefix sum of all
// rows of matrix [,]A and
// store in matrix prefix[]
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
// Update the prefix[,]
if (j == 0)
prefix[i,j] = A[i,j];
else
prefix[i,j] = A[i,j]
+ prefix[i,j - 1];
}
}
// Store the maximum submatrix sum
int maxSum = int.MinValue;
// Iterate for starting column
for (int i = 0; i < c; i++) {
// Iterate for last column
for (int j = i; j < c; j++) {
// To store current array
// elements
List v = new List();
// Traverse every row
for (int k = 0; k < r; k++) {
// Store the sum of the
// kth row
int el = 0;
// Update the prefix
// sum
if (i == 0)
el = prefix[k,j];
else
el = prefix[k,j]
- prefix[k,i - 1];
// Push it in a vector
v.Add(el);
}
// Update the maximum
// overall sum
maxSum = Math.Max(maxSum, kadane(v));
}
}
// Print the answer
Console.Write(maxSum+ "\n");
}
// Driver Code
public static void Main(String[] args)
{
int [,]matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
// Function Call
maxSubmatrixSum(matrix);
}
}
// This code is contributed by 29AjayKumar Javascript
输出：
15时间复杂度： O(N 6 )
辅助空间： O(1)
使用 Kadane 算法的高效方法：可以使用以下观察优化上述方法：
- 修复所需子矩阵的开始和结束列分别说开始和结束。
- 现在，迭代每一行并将从开始到结束列的行总和添加到sumSubmatrix并将其插入到数组中。迭代每一行后，对这个新创建的数组执行 Kadane 算法。如果应用 Kadane 算法获得的总和大于总最大总和，则更新总最大总和。
- 在上述步骤中，可以通过创建包含每行前缀和的大小为N*M的辅助矩阵，在恒定时间内计算从开始到结束列的行和。
请按照以下步骤解决问题：
- 初始化一个变量，比如maxSum为 INT_MIN，以存储最大子数组和。
- 创建一个矩阵prefMatrix[N][M] ，用于存储给定矩阵每一行的前缀数组和。
- 使用i作为行索引和j作为列索引逐行遍历矩阵，并执行以下步骤：
- 如果i 的值为0 ，则设置prefMatrix[i][j] = A[i][j] 。
- 否则，设置prefMatrix[i][j] = prefMatrix[i][j – 1] + A[i][j] 。
- 现在，对于[0, M]范围内子矩阵列的开始和结束索引的所有可能组合，执行以下步骤：
- 初始化一个辅助数组A[]来存储当前子矩阵每一行的最大和。
- 使用prefMatrix求从开始到结束列的总和，如下所示：
- 如果start 的值为正，则将所需的总和S存储为prefMatrix[i][end] – prefMatrix[i][start – 1] 。
- 否则，将S更新为prefMatrix[i][end] 。
- 将S插入数组arr[] 。
- 迭代子矩阵中的所有行后，对数组A[]执行Kadane算法，将最大和maxSum更新为maxSum和本步骤执行Kadane算法得到的值的最大值。
- 完成以上步骤后，打印maxSum的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find maximum continuous
// maximum sum in the array
int kadane(vector v)
{
// Stores current and maximum sum
int currSum = 0;
int maxSum = INT_MIN;
// Traverse the array v
for (int i = 0;
i < (int)v.size(); i++) {
// Add the value of the
// current element
currSum += v[i];
// Update the maximum sum
if (currSum > maxSum) {
maxSum = currSum;
}
if (currSum < 0) {
currSum = 0;
}
}
// Return the maximum sum
return maxSum;
}
// Function to find the maximum
// submatrix sum
void maxSubmatrixSum(
vector > A)
{
// Store the rows and columns
// of the matrix
int r = A.size();
int c = A[0].size();
// Create an auxiliary matrix
int** prefix = new int*[r];
// Traverse the matrix, prefix
// and initialize it will all 0s
for (int i = 0; i < r; i++) {
prefix[i] = new int;
for (int j = 0; j < c; j++) {
prefix[i][j] = 0;
}
}
// Calculate prefix sum of all
// rows of matrix A[][] and
// store in matrix prefix[]
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
// Update the prefix[][]
if (j == 0)
prefix[i][j] = A[i][j];
else
prefix[i][j] = A[i][j]
+ prefix[i][j - 1];
}
}
// Store the maximum submatrix sum
int maxSum = INT_MIN;
// Iterate for starting column
for (int i = 0; i < c; i++) {
// Iterate for last column
for (int j = i; j < c; j++) {
// To store current array
// elements
vector v;
// Traverse every row
for (int k = 0; k < r; k++) {
// Store the sum of the
// kth row
int el = 0;
// Update the prefix
// sum
if (i == 0)
el = prefix[k][j];
else
el = prefix[k][j]
- prefix[k][i - 1];
// Push it in a vector
v.push_back(el);
}
// Update the maximum
// overall sum
maxSum = max(maxSum, kadane(v));
}
}
// Print the answer
cout << maxSum << "\n";
}
// Driver Code
int main()
{
vector > matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
// Function Call
maxSubmatrixSum(matrix);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find maximum continuous
// maximum sum in the array
static int kadane(Vector v)
{
// Stores current and maximum sum
int currSum = 0;
int maxSum = Integer.MIN_VALUE;
// Traverse the array v
for (int i = 0;
i < (int)v.size(); i++)
{
// Add the value of the
// current element
currSum += v.get(i);
// Update the maximum sum
if (currSum > maxSum)
{
maxSum = currSum;
}
if (currSum < 0)
{
currSum = 0;
}
}
// Return the maximum sum
return maxSum;
}
// Function to find the maximum
// submatrix sum
static void maxSubmatrixSum(int [][]A)
{
// Store the rows and columns
// of the matrix
int r = A.length;
int c = A[0].length;
// Create an auxiliary matrix
int [][]prefix = new int[r][];
// Traverse the matrix, prefix
// and initialize it will all 0s
for (int i = 0; i < r; i++) {
prefix[i] = new int;
for (int j = 0; j < c; j++) {
prefix[i][j] = 0;
}
}
// Calculate prefix sum of all
// rows of matrix A[][] and
// store in matrix prefix[]
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
// Update the prefix[][]
if (j == 0)
prefix[i][j] = A[i][j];
else
prefix[i][j] = A[i][j]
+ prefix[i][j - 1];
}
}
// Store the maximum submatrix sum
int maxSum = Integer.MIN_VALUE;
// Iterate for starting column
for (int i = 0; i < c; i++) {
// Iterate for last column
for (int j = i; j < c; j++) {
// To store current array
// elements
Vector v = new Vector();
// Traverse every row
for (int k = 0; k < r; k++) {
// Store the sum of the
// kth row
int el = 0;
// Update the prefix
// sum
if (i == 0)
el = prefix[k][j];
else
el = prefix[k][j]
- prefix[k][i - 1];
// Push it in a vector
v.add(el);
}
// Update the maximum
// overall sum
maxSum = Math.max(maxSum, kadane(v));
}
}
// Print the answer
System.out.print(maxSum+ "\n");
}
// Driver Code
public static void main(String[] args)
{
int [][]matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
// Function Call
maxSubmatrixSum(matrix);
}
}
// This code is contributed by 29AjayKumar
蟒蛇3
# Python3 program for the above approach
import sys
# Function to find maximum continuous
# maximum sum in the array
def kadane(v):
# Stores current and maximum sum
currSum = 0
maxSum = -sys.maxsize - 1
# Traverse the array v
for i in range(len(v)):
# Add the value of the
# current element
currSum += v[i]
# Update the maximum sum
if (currSum > maxSum):
maxSum = currSum
if (currSum < 0):
currSum = 0
# Return the maximum sum
return maxSum
# Function to find the maximum
# submatrix sum
def maxSubmatrixSum(A):
# Store the rows and columns
# of the matrix
r = len(A)
c = len(A[0])
# Create an auxiliary matrix
# Traverse the matrix, prefix
# and initialize it will all 0s
prefix = [[0 for i in range(c)]
for j in range(r)]
# Calculate prefix sum of all
# rows of matrix A[][] and
# store in matrix prefix[]
for i in range(r):
for j in range(c):
# Update the prefix[][]
if (j == 0):
prefix[i][j] = A[i][j]
else:
prefix[i][j] = A[i][j] + prefix[i][j - 1]
# Store the maximum submatrix sum
maxSum = -sys.maxsize - 1
# Iterate for starting column
for i in range(c):
# Iterate for last column
for j in range(i, c):
# To store current array
# elements
v = []
# Traverse every row
for k in range(r):
# Store the sum of the
# kth row
el = 0
# Update the prefix
# sum
if (i == 0):
el = prefix[k][j]
else:
el = prefix[k][j] - prefix[k][i - 1]
# Push it in a vector
v.append(el)
# Update the maximum
# overall sum
maxSum = max(maxSum, kadane(v))
# Print the answer
print(maxSum)
# Driver Code
matrix = [ [ 0, -2, -7, 0 ],
[ 9, 2, -6, 2 ],
[ -4, 1, -4, 1 ],
[ -1, 8, 0, -2 ] ]
# Function Call
maxSubmatrixSum(matrix)
# This code is contributed by rag2127
C＃
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Function to find maximum continuous
// maximum sum in the array
static int kadane(List v)
{
// Stores current and maximum sum
int currSum = 0;
int maxSum = int.MinValue;
// Traverse the array v
for (int i = 0;
i < (int)v.Count; i++)
{
// Add the value of the
// current element
currSum += v[i];
// Update the maximum sum
if (currSum > maxSum)
{
maxSum = currSum;
}
if (currSum < 0)
{
currSum = 0;
}
}
// Return the maximum sum
return maxSum;
}
// Function to find the maximum
// submatrix sum
static void maxSubmatrixSum(int [,]A)
{
// Store the rows and columns
// of the matrix
int r = A.GetLength(0);
int c = A.GetLength(1);
// Create an auxiliary matrix
int [,]prefix = new int[r,c];
// Traverse the matrix, prefix
// and initialize it will all 0s
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
prefix[i,j] = 0;
}
}
// Calculate prefix sum of all
// rows of matrix [,]A and
// store in matrix prefix[]
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
// Update the prefix[,]
if (j == 0)
prefix[i,j] = A[i,j];
else
prefix[i,j] = A[i,j]
+ prefix[i,j - 1];
}
}
// Store the maximum submatrix sum
int maxSum = int.MinValue;
// Iterate for starting column
for (int i = 0; i < c; i++) {
// Iterate for last column
for (int j = i; j < c; j++) {
// To store current array
// elements
List v = new List();
// Traverse every row
for (int k = 0; k < r; k++) {
// Store the sum of the
// kth row
int el = 0;
// Update the prefix
// sum
if (i == 0)
el = prefix[k,j];
else
el = prefix[k,j]
- prefix[k,i - 1];
// Push it in a vector
v.Add(el);
}
// Update the maximum
// overall sum
maxSum = Math.Max(maxSum, kadane(v));
}
}
// Print the answer
Console.Write(maxSum+ "\n");
}
// Driver Code
public static void Main(String[] args)
{
int [,]matrix = { { 0, -2, -7, 0 },
{ 9, 2, -6, 2 },
{ -4, 1, -4, 1 },
{ -1, 8, 0, -2 } };
// Function Call
maxSubmatrixSum(matrix);
}
}
// This code is contributed by 29AjayKumar
Javascript
输出：
15时间复杂度： O(N 3 )
辅助空间： O(N 2 )
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