使用间接递归打印数字 1 到 n
原文:https://www . geesforgeks . org/print-numbers-1-n-using-间接-递归/
给定一个数字 N，我们需要打印从 1 到 N 的数字，不需要直接递归、循环和标签。基本上我们需要插入上面的代码片段，这样它就可以打印从 1 到 N 的数字？
C
#include <stdio.h>
#define N 20;
int main()
{
// Your code goes Here.
}
示例:
Input : 10
Output : 1 2 3 4 5 6 7 8 9 10
Input : 5
Output : 1 2 3 4 5
我们已经在下面的帖子中讨论了解决方案: 用 C++打印 1 到 100，没有循环和递归 不使用循环你怎么打印 1 到 100 的数字？
下面的代码可以打印从 1 到 100 的数字，不需要直接递归、循环和标签。代码使用间接递归。
C++
// C++ program to print from 1 to N using
// indirect recursion/
#include<bits/stdc++.h>
using namespace std;
// We can avoid use of these using references
int N = 20;
int n = 1;
void fun1();
void fun2();
// Prints n, increments n and calls fun1()
void fun1()
{
if (n <= N)
{
cout << n << " ";
n++;
fun2();
}
else
return;
}
// Prints n, increments n and calls fun2()
void fun2()
{
if (n <= N)
{
cout << n << " ";
n++;
fun1();
}
else
return;
}
// Driver Program
int main()
{
fun1();
return 0;
}
// This code is contributed by pankajsharmagfg.
C
// C program to print from 1 to N using
// indirect recursion/
#include<stdio.h>
// We can avoid use of these using references
#define N 20;
int n = 1;
// Prints n, increments n and calls fun1()
void fun1()
{
if (n <= N)
{
printf("%d", n);
n++;
fun2();
}
else
return;
}
// Prints n, increments n and calls fun2()
void fun2()
{
if (n <= N)
{
printf("%d", n);
n++;
fun1();
}
else
return;
}
// Driver Program
int main(void)
{
fun1();
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print from 1 to N using
// indirect recursion
class GFG
{
// We can avoid use of these using references
static final int N = 20;
static int n = 1;
// Prints n, increments n and calls fun1()
static void fun1()
{
if (n <= N)
{
System.out.printf("%d ", n);
n++;
fun2();
}
else
{
return;
}
}
// Prints n, increments n and calls fun2()
static void fun2()
{
if (n <= N)
{
System.out.printf("%d ", n);
n++;
fun1();
}
else
{
return;
}
}
// Driver Program
public static void main(String[] args)
{
fun1();
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python program to prfrom 1 to N using
# indirect recursion
# We can avoid use of these using references
N = 20;
n = 1;
# Prints n, increments n and calls fun1()
def fun1():
global N, n;
if (n <= N):
print(n, end = " ");
n += 1;
fun2();
else:
return;
# Prints n, increments n and calls fun2()
def fun2():
global N, n;
if (n <= N):
print(n, end = " ");
n += 1;
fun1();
else:
return;
# Driver Program
if __name__ == '__main__':
fun1();
# This code is contributed by 29AjayKumar
C
// C# program to print from 1 to N using
// indirect recursion
using System;
class GFG
{
// We can avoid use of these using references
static readonly int N = 20;
static int n = 1;
// Prints n, increments n and calls fun1()
static void fun1()
{
if (n <= N)
{
Console.Write("{0} ", n);
n++;
fun2();
}
else
{
return;
}
}
// Prints n, increments n and calls fun2()
static void fun2()
{
if (n <= N)
{
Console.Write("{0} ", n);
n++;
fun1();
}
else
{
return;
}
}
// Driver Code
public static void Main(String[] args)
{
fun1();
}
}
// This code is contributed by Rajput-Ji
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print
// from 1 to N using
// indirect recursion
// We can avoid use of
// these using references
$N = 20;
$n = 1;
// Prints n, increments
// n and calls fun1()
function fun1()
{
global $N;
global $n;
if ($n <= $N)
{
echo $n, " ";
$n++;
fun2();
}
else
return;
}
// Prints n, increments
// n and calls fun2()
function fun2()
{
global $N;
global $n;
if ($n <= $N)
{
echo $n, " ";
$n++;
fun1();
}
else
return;
}
// Driver Code
fun1();
// This code is contributed
// by m_kit
?>
java 描述语言
<script>
// Javascript program to print from 1 to N using
// indirect recursion
// We can avoid use of these using references
let N = 20;
let n = 1
// Prints n, increments n and calls fun1()
function fun1()
{
if (n <= N)
{
document.write( n+" ");
n++;
fun2();
}
else
{
return;
}
}
// Prints n, increments n and calls fun2()
function fun2()
{
if (n <= N)
{
document.write(n+" ");
n++;
fun1();
}
else
{
return;
}
}
// Driver code
fun1();
// This code is contributed by rag2127
</script>
输出:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
这是如何工作的？: 在上面的程序中，我们只使用了两个函数。一个叫别人，另一个叫前一个，所以间接递归。
练习: 修改上面的程序，使用 N 作为参数，而不是使其全局化。
本文由海得拉巴 Jntuh 工程学院的Umamaheswararao Tumma供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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