使用单循环打印 2D 阵列或矩阵
原文:https://www . geesforgeks . org/print-a-2d-array-or-matrix-use-single-loop/
给定一个 N * M 尺寸的矩阵 mat[][] ，任务是使用单个循环打印矩阵的元素。
示例:
输入: mat[][] = {{1，2，3}，{4，5，6}，{7，8，9 } } T3】输出: 1 2 3 4 5 6 7 8 9
输入: mat[][] = {{7，9}，{10，34}，{12，15 } } T3】输出: 7 9 10 34 12 15
方法:到使用单个循环遍历给定矩阵，观察只有 N * M 元素。因此，想法是使用模数和除法来切换行和列，同时在范围【0，N * M】上迭代单个循环。按照以下步骤解决给定的问题:
- 使用变量 i 在范围【0，N * M】内循环。
- 每次迭代时，求当前行和列的索引分别为行= i / M 和列= i % M 。
- 在上面的步骤中，打印mat【row】【column】的值，得到该索引处矩阵的值。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to print the element
// of 2D matrix using single loop
void print2DMatrix(int arr[][3],
int rows, int columns)
{
// Iterate over the range
// [0, rows*columns]
for(int i = 0; i < rows * columns; i++)
{
// Find row and column index
int row = i / columns;
int col = i % columns;
// Print the element at
// current index
cout << arr[row][col] << " ";
}
}
// Driver Code
int main()
{
// Given matrix mat[][]
int mat[][3] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// Dimensions of the matrix
int N = sizeof(mat) / sizeof(mat[0]);
int M = sizeof(mat[0]) / sizeof(mat[0][0]);
// Function Call
print2DMatrix(mat, N, M);
return 0;
}
// This code is contributed by akhilsaini
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program for the above approach
import java.io.*;
class GFG {
// Function to print the element
// of 2D matrix using single loop
public static void print2DMatrix(
int arr[][], int rows, int columns)
{
// Iterate over the range
// [0, rows*columns]
for (int i = 0;
i < rows * columns; i++) {
// Find row and column index
int row = i / columns;
int col = i % columns;
// Print the element at
// current index
System.out.print(
arr[row][col] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given matrix mat[][]
int[][] mat = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// Dimensions of the matrix
int N = mat.length;
int M = mat[0].length;
// Function Call
print2DMatrix(mat, N, M);
}
}
Python 3
# Python3 program for the above approach
# Function to print the element
# of 2D matrix using single loop
def print2DMatrix(arr, rows, columns):
# Iterate over the range
# [0, rows*columns]
for i in range(0, rows * columns):
# Find row and column index
row = i // columns
col = i % columns
# Print the element at
# current index
print(arr[row][col], end = ' ')
# Driver Code
if __name__ == '__main__':
# Given matrix mat[][]
mat = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ]
# Dimensions of the matrix
N = len(mat)
M = len(mat[0])
# Function Call
print2DMatrix(mat, N, M)
# This code is contributed by akhilsaini
C
// C# program for the above approach
using System;
class GFG{
// Function to print the element
// of 2D matrix using single loop
public static void print2DMatrix(int[, ] arr, int rows,
int columns)
{
// Iterate over the range
// [0, rows*columns]
for(int i = 0; i < rows * columns; i++)
{
// Find row and column index
int row = i / columns;
int col = i % columns;
// Print the element at
// current index
Console.Write(arr[row, col] + " ");
}
}
// Driver Code
public static void Main()
{
// Given matrix mat[][]
int[, ] mat = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// Dimensions of the matrix
int N = mat.GetLength(0);
int M = mat.GetLength(1);
// Function Call
print2DMatrix(mat, N, M);
}
}
// This code is contributed by akhilsaini
java 描述语言
<script>
// JavaScript program for the above approach
// Function to print the element
// of 2D matrix using single loop
function print2DMatrix(arr,rows,columns)
{
// Iterate over the range
// [0, rows*columns]
for(let i = 0; i < rows * columns; i++)
{
// Find row and column index
let row = parseInt(i / columns, 10);
let col = i % columns;
// Print the element at
// current index
document.write(arr[row][col] + " ");
}
}
// Given matrix mat[][]
let mat = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ];
// Dimensions of the matrix
let N = mat.length;
let M = mat[0].length;
// Function Call
print2DMatrix(mat, N, M);
</script>
Output:
1 2 3 4 5 6 7 8 9
时间复杂度: O(N * M) 辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
