最大化矩阵的第K个列的总和

给定两个整数N和K，该任务是最大化K的总和在[1，N ^2]的范围内包括元件的N * N行方向排序的矩阵的^第n列。

例子：

Input: N = 2, K = 2
Output: {{1, 3}, {2, 4}}
Explanations: The possible row-wise sorted matrices are [{{1, 2}, {3, 4}}, {{1, 3}, {2, 4}}, {{1, 4}, {2, 3}}, {{3, 4}, {1, 2}}, {{2, 4}, {1, 3}}, {{2, 3}, {1, 4}} ] 
Out of all the above possible matrices, the matrices [{{1, 3}, {2, 4}}, {{2, 4}, {1, 3}}, {{1, 4}, {2, 3}}, {{2, 3}, {1, 4}}] contains the maximum possible sum of the K^th column. 
Therefore, one of the possible output is {{1, 3}, {2, 4}}.

Input: N = 3, K = 2
Output: {{1, 4, 5}, {2, 6, 7}, {3, 8, 9}}

方法：这里的想法是先用范围[1，N *(K – 1)]中的值填充小于矩阵的^第K列的索引，然后填充大于或等于列的所有元素如下图所示–^第k列由值从区间[+ 1，N ² N *(1 K)]。

请按照以下步骤解决问题：

1. 用[1，N *(K – 1)]范围内的值填充小于K的矩阵的所有列。
2. 然后，用[N *(K – 1)+ 1，N * N]范围内的值填充大于或等于K的矩阵列。
3. 最后，打印矩阵。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to maximize the Kth column sum
int** findMatrix(int N, int K)
{
 
    // Store all the elements of the
    // resultant matrix of size N*N
    int** mat = (int**)malloc(
        N * sizeof(int*));
 
    for (int i = 0; i < N; ++i) {
        mat[i] = (int*)malloc(
            N * sizeof(int));
    }
 
    // Store value of each
    // elements of the matrix
    int element = 1;
 
    // Fill all the columns < K
    for (int i = 0; i < N; ++i) {
 
        for (int j = 0; j < K - 1; ++j) {
            mat[i][j] = element++;
        }
    }
 
    // Fill all the columns >= K
    for (int i = 0; i < N; ++i) {
 
        for (int j = K - 1; j < N; ++j) {
            mat[i][j] = element++;
        }
    }
 
    return mat;
}
 
// Function to print the matrix
void printMatrix(int** mat, int N)
{
 
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
 
    int N = 3, K = 2;
    int** mat = findMatrix(N, K);
 
    printMatrix(mat, N);
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to maximize the Kth column sum
static int [][]findMatrix(int N, int K)
{
 
    // Store all the elements of the
    // resultant matrix of size N*N
    int [][]mat = new int[N][N];
 
    // Store value of each
    // elements of the matrix
    int element = 1;
 
    // Fill all the columns < K
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < K - 1; ++j)
        {
            mat[i][j] = element++;
        }
    }
 
    // Fill all the columns >= K
    for(int i = 0; i < N; ++i)
    {
        for(int j = K - 1; j < N; ++j)
        {
            mat[i][j] = element++;
        }
    }
    return mat;
}
 
// Function to print the matrix
static void printMatrix(int [][]mat, int N)
{
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < N; ++j)
        {
            System.out.print(mat[i][j] + " ");
        }
        System.out.println();
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 3, K = 2;
    int [][]mat = findMatrix(N, K);
 
    printMatrix(mat, N);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to maximize the Kth
# column sum
def findMatrix(N, K):
     
    # Store all the elements of the
    # resultant matrix of size N*N
    mat = [[0 for i in range(N)]
              for j in range(N)];
 
    # Store value of each
    # elements of the matrix
    element = 0;
 
    # Fill all the columns < K
    for i in range(0, N):
        for j in range(0, K - 1):
            element += 1;
            mat[i][j] = element;
 
    # Fill all the columns >= K
    for i in range(0, N):
        for j in range(K - 1, N):
            element += 1;
            mat[i][j] = element;
 
    return mat;
 
# Function to prthe matrix
def printMatrix(mat, N):
     
    for i in range(0, N):
        for j in range(0, N):
            print(mat[i][j], end = " ");
 
        print();
 
# Driver Code
if __name__ == '__main__':
     
    N = 3; K = 2;
    mat = findMatrix(N, K);
 
    printMatrix(mat, N);
 
# This code is contributed by Amit Katiyar

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to maximize the Kth column sum
static int [,]findMatrix(int N, int K)
{
 
    // Store all the elements of the
    // resultant matrix of size N*N
    int [,]mat = new int[N, N];
 
    // Store value of each
    // elements of the matrix
    int element = 1;
 
    // Fill all the columns < K
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < K - 1; ++j)
        {
            mat[i, j] = element++;
        }
    }
 
    // Fill all the columns >= K
    for(int i = 0; i < N; ++i)
    {
        for(int j = K - 1; j < N; ++j)
        {
            mat[i, j] = element++;
        }
    }
    return mat;
}
 
// Function to print the matrix
static void printMatrix(int [,]mat, int N)
{
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < N; ++j)
        {
            Console.Write(mat[i, j] + " ");
        }
        Console.WriteLine();
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 3, K = 2;
    int [,]mat = findMatrix(N, K);
 
    printMatrix(mat, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to maximize the Kth column sum
function findMatrix(N, K)
{
     
    // Store all the elements of the
    // resultant matrix of size N*N
    let mat =  new Array(N);
 
    // Loop to create 2D array using 1D array
    for(var i = 0; i < mat.length; i++)
    {
        mat[i] = new Array(2);
    }
   
    // Store value of each
    // elements of the matrix
    let element = 1;
   
    // Fill all the columns < K
    for(let i = 0; i < N; ++i)
    {
        for(let j = 0; j < K - 1; ++j)
        {
            mat[i][j] = element++;
        }
    }
   
    // Fill all the columns >= K
    for(let i = 0; i < N; ++i)
    {
        for(let j = K - 1; j < N; ++j)
        {
            mat[i][j] = element++;
        }
    }
    return mat;
}
   
// Function to prlet the matrix
function prletMatrix(mat, N)
{
    for(let i = 0; i < N; ++i)
    {
        for(let j = 0; j < N; ++j)
        {
            document.write(mat[i][j] + " ");
        }
        document.write("<br/>");
    }
}
 
// Driver code
let N = 3, K = 2;
let mat = findMatrix(N, K);
 
prletMatrix(mat, N);
 
// This code is contributed by souravghosh0416
 

输出：

1 4 5 
2 6 7 
3 8 9

时间复杂度： O(N ² )
辅助空间： O(N ² )