程序打印给定的Z模式

给定整数N ，任务是打印字母Z模式，如下所示：

1 2 3 * * * N
          *
        *
      *
    3
  2
1 2 3 * * * N

例子：

Input: N = 5
Output: 
1 2 3 4 5 
      4
    3
  2
1 2 3 4 5

Input: N = 4
Output: 
1 2 3 4
    3
  2
1 2 3 4        

方法：

• 用1到N的数字打印第一行。
• 然后从第二行到第(N-1)行，打印2 *(N – index – 1)乘以空格，后跟结束元素，即index – 1 。
• 用1到N的数字打印最后一行。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to print the desired
// Alphabet Z Pattern
void alphabetPattern(int N)
{
 
    int index, side_index, size;
 
    // Declaring the values of Right,
    // Left and Diagonal values
    int Top = 1, Bottom = 1, Diagonal = N - 1;
 
    // Loop for printing the first row
    for (index = 0; index < N; index++)
        cout << Top++ << " ";
 
    cout << endl;
 
    // Main Loop for the rows from (2 to n-1)
    for (index = 1; index < N - 1; index++) {
 
        // Spaces for the diagonals
        for (side_index = 0; side_index < 2 * (N - index - 1);
             side_index++)
            cout << " ";
 
        // Printing the diagonal values
        cout << Diagonal--;
 
        cout << endl;
    }
 
    // Loop for printing the last row
    for (index = 0; index < N; index++)
        cout << Bottom++ << " ";
}
 
// Driver Code
int main()
{
    // Number of rows
    int N = 5;
 
    alphabetPattern(N);
 
    return 0;
}

C

// C implementation of the approach
#include 
 
// Function to print the desired
// Alphabet Z Pattern
void alphabet_Z_Pattern(int N)
{
    int index, side_index, size;
 
    // Declaring the values of Right,
    // Left and Diagonal values
    int Top = 1, Bottom = 1, Diagonal = N - 1;
 
    // Loop for printing the first row
    for (index = 0; index < N; index++)
        printf("%d ", Top++);
 
    printf("\n");
 
    // Main Loop for the rows from (2 to n-1)
    for (index = 1; index < N - 1; index++) {
 
        // Spaces for the diagonals
        for (side_index = 0; side_index < 2 * (N - index - 1);
             side_index++)
            printf(" ");
 
        // Printing the diagonal values
        printf("%d", Diagonal--);
 
        printf("\n");
    }
 
    // Loop for printing the last row
    for (index = 0; index < N; index++)
        printf("%d ", Bottom++);
}
 
// Driver Code
int main()
{
    // Size of the Pattern
    int N = 5;
 
    alphabet_Z_Pattern(N);
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
// Function to print the desired
// Alphabet Z Pattern
static void alphabetPattern(int N)
{
 
    int index, side_index;
 
    // Declaring the values of Right,
    // Left and Diagonal values
    int Top = 1, Bottom = 1, Diagonal = N - 1;
 
    // Loop for printing the first row
    for (index = 0; index < N; index++)
        System.out.print(Top++ + " ");
 
    System.out.println();
 
    // Main Loop for the rows from (2 to n-1)
    for (index = 1; index < N - 1; index++)
    {
 
        // Spaces for the diagonals
        for (side_index = 0;
             side_index < 2 * (N - index - 1);
            side_index++)
            System.out.print(" ");
 
        // Printing the diagonal values
        System.out.print(Diagonal--);
 
        System.out.println();
    }
 
    // Loop for printing the last row
    for (index = 0; index < N; index++)
        System.out.print(Bottom++ + " ");
}
 
// Driver Code
public static void main(String args[])
{
    // Number of rows
    int N = 5;
 
    alphabetPattern(N);
}
}
 
// This code is contributed
// by Akanksha Rai

Python3

# Python 3 implementation of the approach
 
# Function to print the desired
# Alphabet Z Pattern
def alphabetPattern(N):
 
    # Declaring the values of Right,
    # Left and Diagonal values
    Top, Bottom, Diagonal = 1, 1, N - 1
 
    # Loop for printing the first row
    for index in range(N):
        print(Top, end = ' ')
        Top += 1
    print()
 
    # Main Loop for the rows from (2 to n-1)
    for index in range(1, N - 1):
 
        # Spaces for the diagonals
        for side_index in range(2 * (N - index - 1)):
            print(' ', end = '')
 
        # Printing the diagonal values
        print(Diagonal, end = '')
        Diagonal -= 1
        print()
 
    # Loop for printing the last row
    for index in range(N):
        print(Bottom, end = ' ')
        Bottom += 1
 
# Driver Code
 
# Number of rows
N = 5
alphabetPattern(N)
 
# This code is contributed
# by SamyuktaSHegde

C#

// C# implementation of the approach
using System;
 
class GFG
{
// Function to print the desired
// Alphabet Z Pattern
static void alphabetPattern(int N)
{
 
    int index, side_index;
 
    // Declaring the values of Right,
    // Left and Diagonal values
    int Top = 1, Bottom = 1, Diagonal = N - 1;
 
    // Loop for printing the first row
    for (index = 0; index < N; index++)
        Console.Write(Top++ + " ");
 
    Console.WriteLine();
 
    // Main Loop for the rows from (2 to n-1)
    for (index = 1; index < N - 1; index++)
    {
 
        // Spaces for the diagonals
        for (side_index = 0; side_index < 2 * (N - index - 1);
            side_index++)
            Console.Write(" ");
 
        // Printing the diagonal values
        Console.Write(Diagonal--);
 
        Console.WriteLine();
    }
 
    // Loop for printing the last row
    for (index = 0; index < N; index++)
        Console.Write(Bottom++ + " ");
}
 
// Driver Code
public static void Main()
{
    // Number of rows
    int N = 5;
 
    alphabetPattern(N);
}
}
 
// This code is contributed
// by Akanksha Rai

---

Javascript

 
      // JavaScript implementation of the approach
 
      // Function to print the desired
      // Alphabet Z Pattern
      function alphabetPattern(N)
      {
        var index, side_index, size;
 
        // Declaring the values of Right,
        // Left and Diagonal values
        var Top = 1,
          Bottom = 1,
          Diagonal = N - 1;
 
        // Loop for printing the first row
        for (index = 0; index < N; index++) {
          document.write(Top + "  ");
          Top++;
        }
 
        document.write("<br>");
 
        // Main Loop for the rows from (2 to n-1)
        for (index = 1; index < N - 1; index++) {
          // Spaces for the diagonals
          for (side_index = 0; side_index <
           2 * (N - index - 1); side_index++)
            document.write("  ");
 
          // Printing the diagonal values
          document.write(Diagonal);
          Diagonal--;
 
          document.write("<br>");
        }
 
        // Loop for printing the last row
        for (index = 0; index < N; index++) {
          document.write(Bottom + "  ");
          Bottom++;
        }
      }
 
      // Driver Code
      // Number of rows
      var N = 5;
      alphabetPattern(N);
       
 

输出：

1 2 3 4 5 
      4
    3
  2
1 2 3 4 5

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