原文:
给定一个mat【】【】，以波形打印。
输入: mat[][] = {{ 1，2，3，4} { 5，6，7，8} { 9，10，11，12} {13，14，15，16} {17，18，19，20}} 输出:1 5 9 13 17 18 14 10 6 2 3 11 15 19 20 16 12 8 4【t4
输入: mat[][] = {{1，9，4，10} { 3，6，90，11} { 2，30，85，72} { 6，31，99，15 } } t6】输出: 1 3 2 6 31 30 6 9 4 90 85 99 15 72 11 10
方法:这个问题是基于实现的，并且有一个类似于文章中讨论的方法。要获得给定的所需波形，首先，向下打印矩阵第一列的元素，然后向上打印 2 和列的元素，然后向下打印第三列的元素，以此类推。
下面是上述方法的实现:
c
// c program for above approach
#include
using namespace std;
#define r 5
#define c 4
// function to print wave
// form for a given matrix
void waveprint(int m, int n, int arr[r][c])
{
// loop to traverse matrix
for (int j = 0; j < n; j ) {
// if the current column
// is even indexed, print
// it in forward order
if (j % 2 == 0) {
for (int i = 0; i < m; i ) {
cout << arr[i][j] << " ";
}
}
// if the current column
// is odd indexed, print
// it in reverse order
else {
for (int i = m - 1; i >= 0; i--) {
cout << arr[i][j] << " ";
}
}
}
}
// driver code
int main()
{
int arr[r][c] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 },
{ 17, 18, 19, 20 } };
waveprint(r, c, arr);
return 0;
}
c
// c program for above approach
#include
#define r 5
#define c 4
// function to print wave
// form for a given matrix
void waveprint(int m, int n, int arr[r][c])
{
// loop to traverse matrix
for (int j = 0; j < n; j ) {
// if the current column
// is even indexed, print
// it in forward order
if (j % 2 == 0) {
for (int i = 0; i < m; i ) {
printf("%d ", arr[i][j]);
}
}
// if the current column
// is odd indexed, print
// it in reverse order
else {
for (int i = m - 1; i >= 0; i--) {
printf("%d ", arr[i][j]);
}
}
}
}
// driver code
int main()
{
int arr[r][c] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 },
{ 13, 14, 15, 16 } };
waveprint(r, c, arr);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for above approach
import java.io.*;
class gfg {
public static int r = 5;
public static int c = 4;
// function to print wave
// form for a given matrix
public static void waveprint(int m, int n, int[][] arr)
{
// loop to traverse matrix
for (int j = 0; j < n; j ) {
// if the current column
// is even indexed, print
// it in forward order
if (j % 2 == 0) {
for (int i = 0; i < m; i ) {
system.out.print(arr[i][j] " ");
}
}
// if the current column
// is odd indexed, print
// it in reverse order
else {
for (int i = m - 1; i >= 0; i--) {
system.out.print(arr[i][j] " ");
}
}
}
}
// driver code
public static void main (string[] args)
{
int[][] arr = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 },
{ 17, 18, 19, 20 } };
waveprint(r, c, arr);
}
}
// this code is contributed by shubham singh
python 3
# python code for the above approach
r = 5
c = 4
# function to print wave
# form for a given matrix
def waveprint(m, n, arr):
# loop to traverse matrix
for j in range(n):
# if the current column
# is even indexed, print
# it in forward order
if (j % 2 == 0):
for i in range(m):
print(arr[i][j], end= " ")
# if the current column
# is odd indexed, print
# it in reverse order
else:
for i in range(m - 1, -1, -1):
print(arr[i][j], end= " ")
# driver code
arr = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16], [17, 18, 19, 20]]
waveprint(r, c, arr)
# this code is contributed by gfgking
c
// c# program for above approach
using system;
class gfg{
public static int r = 5;
public static int c = 4;
// function to print wave
// form for a given matrix
public static void waveprint(int m, int n, int[,] arr)
{
// loop to traverse matrix
for(int j = 0; j < n; j )
{
// if the current column
// is even indexed, print
// it in forward order
if (j % 2 == 0)
{
for(int i = 0; i < m; i )
{
console.write(arr[i, j] " ");
}
}
// if the current column
// is odd indexed, print
// it in reverse order
else
{
for(int i = m - 1; i >= 0; i--)
{
console.write(arr[i, j] " ");
}
}
}
}
// driver code
public static void main ()
{
int[,] arr = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 },
{ 17, 18, 19, 20 } };
waveprint(r, c, arr);
}
}
// this code is contributed by gfgking
java 描述语言
output
1 5 9 13 17 18 14 10 6 2 3 7 11 15 19 20 16 12 8 4
时间复杂度:o(n2) 辅助空间: o(1)
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