给定整数N(其为行数)，任务是绘制双箭头形状的数字图案。
先决条件：该模式是增长和收缩类型的模式，因此需要执行循环的基础知识才能理解主题和任何语言的代码。几何形状可以可视化为- 
例子：
Input: R = 7
Output:
1
2 1 1 2
3 2 1 1 2 3
4 3 2 1 1 2 3 4
3 2 1 1 2 3
2 1 1 2
1
Input: R = 9
Output:
1
2 1 1 2
3 2 1 1 2 3
4 3 2 1 1 2 3 4
5 4 3 2 1 1 2 3 4 5
4 3 2 1 1 2 3 4
3 2 1 1 2 3
2 1 1 2
1
方法：
- 在给定的示例中，N = 7，ROWS的数目为7。
- 垂直地，图案增长到ROW = N / 2，然后收缩。
- 第1行有4个“(空格)字符，然后是一个值。
- 在每个连续的行中，SPACE字符的数量减少，而NUMERALS个计数增加。
- 另外，请注意，每行中放置的数字的第1个值与该行的数目相同。
- 同样，水平模式具有NUMERALS，然后是SPACES和NUMERALS。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the requried pattern
void drawPattern(int N)
{
int n = N;
int row = 1;
// 'nst' is the number of values
int nst = 1;
// 'nsp' is the number of spaces
int nsp1 = n - 1;
int nsp2 = -1;
int val1 = row;
int val2 = 1;
while (row <= n) {
// Here spaces are printed
// 'csp' is the count of spaces
int csp1 = 1;
while (csp1 <= nsp1) {
cout << " "
<< " ";
csp1 = csp1 + 1;
}
// Now, values are printed
// 'cst' is the count of stars
int cst1 = 1;
while (cst1 <= nst) {
cout << val1 << " ";
val1 = val1 - 1;
cst1 = cst1 + 1;
}
// Again spaces have to be printed
int csp2 = 1;
while (csp2 <= nsp2) {
cout << " "
<< " ";
csp2 = csp2 + 1;
}
// Again values have to be printed
if (row != 1 && row != n) {
int cst2 = 1;
while (cst2 <= nst) {
cout << val2 << " ";
val2 = val2 + 1;
cst2 = cst2 + 1;
}
}
cout << endl;
// Move to the next row
if (row <= n / 2) {
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
}
else {
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
}
row = row + 1;
}
}
// Driver code
int main()
{
// Number of rows
int N = 7;
drawPattern(N);
return 0;
} Java
// Java implementation of the approach
class GFG {
// Function to print the requried pattern
static void drawPattern(int N)
{
int n = N;
int row = 1;
// 'nst' is the number of values
int nst = 1;
// 'nsp' is the number of spaces
int nsp1 = n - 1;
int nsp2 = -1;
int val1 = row;
int val2 = 1;
while (row <= n) {
// Here spaces are printed
// 'csp' is the count of spaces
int csp1 = 1;
while (csp1 <= nsp1) {
System.out.print(" ");
csp1 = csp1 + 1;
}
// Now, values are printed
// 'cst' is the count of stars
int cst1 = 1;
while (cst1 <= nst) {
System.out.print(val1 + " ");
val1 = val1 - 1;
cst1 = cst1 + 1;
}
// Again spaces have to be printed
int csp2 = 1;
while (csp2 <= nsp2) {
System.out.print(" ");
csp2 = csp2 + 1;
}
// Again values have to be printed
if (row != 1 && row != n) {
int cst2 = 1;
while (cst2 <= nst) {
System.out.print(val2 + " ");
val2 = val2 + 1;
cst2 = cst2 + 1;
}
}
System.out.println();
// Move to the next row
if (row <= n / 2) {
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
}
else {
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
}
row = row + 1;
}
}
// Driver code
public static void main(String args[])
{
// Number of rows
int N = 7;
drawPattern(N);
}
}Python3
# Python3 implementation of the approach
# Function to print the requried pattern
def drawPattern(N) :
n = N;
row = 1;
# 'nst' is the number of values
nst = 1;
# 'nsp' is the number of spaces
nsp1 = n - 1;
nsp2 = -1;
val1 = row;
val2 = 1;
while (row <= n) :
# Here spaces are printed
# 'csp' is the count of spaces
csp1 = 1;
while (csp1 <= nsp1) :
print(" ",end= " ");
csp1 = csp1 + 1;
# Now, values are printed
# 'cst' is the count of stars
cst1 = 1;
while (cst1 <= nst) :
print(val1,end = " ");
val1 = val1 - 1;
cst1 = cst1 + 1;
# Again spaces have to be printed
csp2 = 1;
while (csp2 <= nsp2) :
print(" ",end = " ");
csp2 = csp2 + 1;
# Again values have to be printed
if (row != 1 and row != n) :
cst2 = 1;
while (cst2 <= nst) :
print(val2,end = " ");
val2 = val2 + 1;
cst2 = cst2 + 1;
print()
# Move to the next row
if (row <= n // 2) :
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
else :
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
row = row + 1;
# Driver code
if __name__ == "__main__" :
# Number of rows
N = 7;
drawPattern(N);
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the requried pattern
static void drawPattern(int N)
{
int n = N;
int row = 1;
// 'nst' is the number of values
int nst = 1;
// 'nsp' is the number of spaces
int nsp1 = n - 1;
int nsp2 = -1;
int val1 = row;
int val2 = 1;
while (row <= n)
{
// Here spaces are printed
// 'csp' is the count of spaces
int csp1 = 1;
while (csp1 <= nsp1)
{
Console.Write(" ");
csp1 = csp1 + 1;
}
// Now, values are printed
// 'cst' is the count of stars
int cst1 = 1;
while (cst1 <= nst)
{
Console.Write(val1 + " ");
val1 = val1 - 1;
cst1 = cst1 + 1;
}
// Again spaces have to be printed
int csp2 = 1;
while (csp2 <= nsp2)
{
Console.Write(" ");
csp2 = csp2 + 1;
}
// Again values have to be printed
if (row != 1 && row != n)
{
int cst2 = 1;
while (cst2 <= nst)
{
Console.Write(val2 + " ");
val2 = val2 + 1;
cst2 = cst2 + 1;
}
}
Console.WriteLine();
// Move to the next row
if (row <= n / 2)
{
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
}
else
{
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
}
row = row + 1;
}
}
// Driver code
public static void Main()
{
// Number of rows
int N = 7;
drawPattern(N);
}
}
// This code is contributed by AnkitRai01输出：
1
2 1 1 2
3 2 1 1 2 3
4 3 2 1 1 2 3 4
3 2 1 1 2 3
2 1 1 2
1
时间复杂度： O(N 2 )
空间复杂度： O(1)