使用递归打印给定范围内的偶数和奇数

给定两个整数L和R ，任务是使用递归打印从L到R的所有偶数和奇数。

例子：

Input: L = 1, R = 10
Output: 
Even numbers: 2 4 6 8 10
Odd numbers: 1 3 5 7 9

Input: L = 10, R = 25 
Output: 
Even numbers:10 12 14 16 18 20 22 24 
Odd numbers:11 13 15 17 19 21 23 25

方法：按照以下步骤使用递归解决问题：

• 遍历范围[R，L] 。
• 使用以下递归关系，使用递归打印范围中的奇数元素：

Odd(L, R) = R % 2 == 1? Odd(L, R – 2) : Odd(L, R – 1) 

• 使用以下递归关系使用递归打印范围中的偶数元素：

Even(L, R) = R % 2 == 0 ? Even(L, R – 2) : Even(L, R – 1) 

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to print all the
// even numbers from L to R
void Even(int L, int R)
{
 
    // Base case
    if (R < L) {
        return;
    }
 
    // Recurrence relation
    R % 2 == 0 ? Even(L, R - 2)
               : Even(L, R - 1);
 
    // Check if R is even
    if (R % 2 == 0) {
        cout << R << " ";
    }
}
 
// Function to print all the
// odd numbers from L to R
void Odd(int L, int R)
{
 
    // Base case
    if (R < L) {
        return;
    }
 
    // Recurrence relation
    R % 2 == 1 ? Odd(L, R - 2)
               : Odd(L, R - 1);
 
    // Check if R is even
    if (R % 2 == 1) {
        cout << R << " ";
    }
}
 
// Driver Code
int main()
{
    int L = 10, R = 25;
    cout << "Even numbers:";
 
    // Print all the
    // even numbers
    Even(L, R);
    cout << endl;
 
    // Print all the
    // odd numbers
    cout << "Odd numbers:";
    Odd(L, R);
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to print
// all the even numbers
// from L to R
static void Even(int L,
                 int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 0 )
    Even(L, R - 2);
  else
    Even(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 0)
  {
    System.out.print(R + " ");
  }
}
 
// Function to print
// all the odd numbers
// from L to R
static void Odd(int L,
                int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 1 )
    Odd(L, R - 2);
  else
    Odd(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 1)
  {
    System.out.print(R + " ");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  int L = 10, R = 25;
  System.out.print("Even numbers:");
 
  // Print all the
  // even numbers
  Even(L, R);
  System.out.println();
 
  // Print all the
  // odd numbers
  System.out.print("Odd numbers:");
  Odd(L, R);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement
# the above approach
 
# Function to print all the
# even numbers from L to R
def Even(L, R):
     
    # Base case
    if (R < L):
        return
 
    # Recurrence relation
    if (R % 2 == 0):
        Even(L, R - 2)
    else:
        Even(L, R - 1)
 
    # Check if R is even
    if (R % 2 == 0):
        print(R, end = " ")
 
# Function to print all the
# odd numbers from L to R
def Odd(L, R):
     
    # Base case
    if (R < L):
        return
 
    # Recurrence relation
    if (R % 2 == 1):
        Odd(L, R - 2)
    else:
        Odd(L, R - 1)
 
    # Check if R is even
    if (R % 2 == 1):
        print(R, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    L = 10
    R = 25
     
    print("Even numbers:")
 
    # Print all the
    # even numbers
    Even(L, R)
    print()
 
    # Print all the
    # odd numbers
    print("Odd numbers:")
    Odd(L, R)
 
# This code is contributed by Amit Katiyar

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to print
// all the even numbers
// from L to R
static void Even(int L,
                 int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 0 )
    Even(L, R - 2);
  else
    Even(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 0)
  {
    Console.Write(R + " ");
  }
}
 
// Function to print
// all the odd numbers
// from L to R
static void Odd(int L,
                int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 1 )
    Odd(L, R - 2);
  else
    Odd(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 1)
  {
    Console.Write(R + " ");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  int L = 10, R = 25;
  Console.Write("Even numbers:");
 
  // Print all the
  // even numbers
  Even(L, R);
  Console.WriteLine();
 
  // Print all the
  // odd numbers
  Console.Write("Odd numbers:");
  Odd(L, R);
}
}
 
// This code is contributed by 29AjayKumar

输出：

Even numbers:10 12 14 16 18 20 22 24 
Odd numbers:11 13 15 17 19 21 23 25

时间复杂度： O(N)
辅助空间： O(1)