将一个矩阵转换成另一个给定维度的矩阵
原文:https://www . geeksforgeeks . org/convert-a-matrix-to-另一个给定维度的矩阵/
给定一个大小为 N * M 的矩阵、 mat[][] 和两个正整数 A 和 B ,任务是根据给定的矩阵元素构造一个大小为 A * B 的矩阵。如果存在多个解决方案,则打印其中任何一个。否则,打印 -1 。
输入: mat[][] = { { 1,2,3,4,5,6 } },A = 2,B = 3 输出: { { 1,2,3 },{ 4,5,6 } } 解释 : 由于矩阵的大小{ { 1,2,3 },{ 4,5,6 } }是 A * B(2 * 3)。 因此,需要的输出是{ { 1,2,3 },{ 4,5,6 } }。
输入: mat[][] = { {1,2},{ 3,4 },{ 5,6 } },A = 1,B = 6 输出: { { 1,2,3,4,5,6 } }
方法:按照以下步骤解决问题:
- 初始化一个大小为 A * B 的矩阵,比如说 res[][] 。
- 遍历矩阵, mat[][] 并将矩阵的每个元素插入矩阵, res[][] 。
- 最后,打印矩阵 res[][] 。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to construct a matrix of size
// A * B from the given matrix elements
void ConstMatrix(int* mat, int N, int M,
int A, int B)
{
if (N * M != A * B)
return;
int idx = 0;
// Initialize a new matrix
int res[A][B];
// Traverse the matrix, mat[][]
for(int i = 0; i < N; i++)
{
for(int j = 0; j < M; j++)
{
res[idx / B][idx % B] = *((mat + i * M) + j);
// Update idx
idx++;
}
}
// Print the resultant matrix
for(int i = 0; i < A; i++)
{
for(int j = 0; j < B; j++)
cout << res[i][j] << " ";
cout << "\n";
}
}
// Driver Code
int main()
{
int mat[][6] = { { 1, 2, 3, 4, 5, 6 } };
int A = 2;
int B = 3;
int N = sizeof(mat) / sizeof(mat[0]);
int M = sizeof(mat[0]) / sizeof(int);
ConstMatrix((int*)mat, N, M, A, B);
return 0;
}
// This code is contributed by subhammahato348
Java 语言(一种计算机语言,尤用于创建网站)
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to cona matrix of size
// A * B from the given matrix elements
static void ConstMatrix(int[][] mat, int N, int M,
int A, int B)
{
if (N * M != A * B)
return;
int idx = 0;
// Initialize a new matrix
int [][]res = new int[A][B];
// Traverse the matrix, mat[][]
for(int i = 0; i < N; i++)
{
for(int j = 0; j < M; j++)
{
res[idx / B][idx % B] = mat[i][j];
// Update idx
idx++;
}
}
// Print the resultant matrix
for(int i = 0; i < A; i++)
{
for(int j = 0; j < B; j++)
System.out.print(res[i][j] + " ");
System.out.print("\n");
}
}
// Driver Code
public static void main(String[] args)
{
int mat[][] = { { 1, 2, 3, 4, 5, 6 } };
int A = 2;
int B = 3;
int N = mat.length;
int M = mat[0].length;
ConstMatrix(mat, N, M, A, B);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program to implement
# the above approach
# Function to construct a matrix of size A * B
# from the given matrix elements
def ConstMatrix(mat, N, M, A, B):
if (N * M != A * B):
return -1
idx = 0;
# Initialize a new matrix
res = [[0 for i in range(B)] for i in range(A)]
# Traverse the matrix, mat[][]
for i in range(N):
for j in range(M):
res[idx//B][idx % B] = mat[i][j];
# Update idx
idx += 1
# Print the resultant matrix
for i in range(A):
for j in range(B):
print(res[i][j], end = " ")
print()
# Driver Code
if __name__ == '__main__':
mat = [ [ 1, 2, 3, 4, 5, 6 ] ]
A = 2
B = 3
N = len(mat)
M = len(mat[0])
ConstMatrix(mat, N, M, A, B)
C
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to cona matrix of size
// A * B from the given matrix elements
static void ConstMatrix(int[,] mat, int N, int M,
int A, int B)
{
if (N * M != A * B)
return;
int idx = 0;
// Initialize a new matrix
int [,]res = new int[A,B];
// Traverse the matrix, [,]mat
for(int i = 0; i < N; i++)
{
for(int j = 0; j < M; j++)
{
res[idx / B, idx % B] = mat[i, j];
// Update idx
idx++;
}
}
// Print the resultant matrix
for(int i = 0; i < A; i++)
{
for(int j = 0; j < B; j++)
Console.Write(res[i, j] + " ");
Console.Write("\n");
}
}
// Driver Code
public static void Main(String[] args)
{
int [,]mat = {{ 1, 2, 3, 4, 5, 6 }};
int A = 2;
int B = 3;
int N = mat.GetLength(0);
int M = mat.GetLength(1);
ConstMatrix(mat, N, M, A, B);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// javascript program of the above approach
// Function to cona matrix of size
// A * B from the given matrix elements
function ConstMatrix(mat, N, M,
A, B)
{
if (N * M != A * B)
return;
let idx = 0;
// Initialize a new matrix
let res = new Array(A);
// Loop to create 2D array using 1D array
for (var i = 0; i < res.length; i++) {
res[i] = new Array(2);
}
// Traverse the matrix, mat[][]
for(let i = 0; i < N; i++)
{
for(let j = 0; j < M; j++)
{
res[Math.floor(idx / B)][idx % B] = mat[i][j];
// Update idx
idx++;
}
}
// Prlet the resultant matrix
for(let i = 0; i < A; i++)
{
for(let j = 0; j < B; j++)
document.write(res[i][j] + " ");
document.write("<br/>");
}
}
// Driver Code
let mat = [[ 1, 2, 3, 4, 5, 6 ]];
let A = 2;
let B = 3;
let N = mat.length;
let M = mat[0].length;
ConstMatrix(mat, N, M, A, B);
// This code is contributed by chinmoy1997pal.
</script>
Output:
1 2 3
4 5 6
时间复杂度: O(N * M) 辅助空间: O(N * M)
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