递归练习题|第 3 集
原文:https://www . geesforgeks . org/practice-questions-for-recursion-set-3/
解释下面递归函数的功能。
问题 1
C++
void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n - 1);
for (i = 0; i < n; i++)
cout << " * ";
}
// This code is contributed by shubhamsingh10
C
void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n-1);
for (i = 0; i < n; i++)
printf(" * ");
}
Java 语言(一种计算机语言，尤用于创建网站)
static void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n - 1);
for (i = 0; i < n; i++)
System.out.print(" * ");
}
// This code is contributed by shubhamsingh10
Python 3
def fun1(n):
i = 0
if (n > 1):
fun1(n - 1)
for i in range(n):
print(" * ",end="")
# This code is contributed by shubhamsingh10
C
static void fun1(int n)
{
int i = 0;
if (n > 1)
fun1(n-1);
for (i = 0; i < n; i++)
Console.Write(" * ");
}
// This code is contributed by shubhamsingh10
java 描述语言
<script>
function fun1(n)
{
let i = 0;
if (n > 1)
fun1(n - 1);
for(i = 0; i < n; i++)
document.write(" * ");
}
// This code is contributed by gottumukkalabobby
</script>
答:打印的星星总数等于 1 + 2 + …。(n-2) + (n-1) + n，即 n(n+1)/2。
问题 2
C++
#define LIMIT 1000
void fun2(int n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
cout << n <<" ";
fun2(2*n);
cout << n <<" ";
}
// This code is contributed by shubhamsingh10
C
#define LIMIT 1000
void fun2(int n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
printf("%d ", n);
fun2(2*n);
printf("%d ", n);
}
Java 语言(一种计算机语言，尤用于创建网站)
int LIMIT = 1000;
void fun2(int n)
{
if (n <= 0) return;
if (n > LIMIT) return;
System.out.print(String.format("%d ", n));
fun2(2 * n);
System.out.print(String.format("%d ", n));
}
Python 3
LIMIT = 1000
def fun2(n):
if (n <= 0):
return
if (n > LIMIT):
return
print(n, end=" ")
fun2(2 * n)
print(n, end=" ")
# This code is contributed by shubhamsingh10
C
int LIMIT = 1000
void fun2(int n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
Console.Write(n+" ");
fun2(2*n);
Console.Write(n+" ");
}
// This code is contributed by Shubhamsingh10
java 描述语言
<script>
let LIMIT = 1000;
function fun2(n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
document.write(n + " "));
fun2(2 * n);
document.write(n + " "));
}
// This code is contributed by gottumukkalabobby
</script>
答:对于正 n，fun2(n)打印 n，2n，4n，8n …的值，而该值小于 LIMIT。以递增顺序打印数值后，它会以相反的顺序再次打印相同的数字。例如，fun2(100)打印 100，200，400，800，800，400，200，100。 如果 n 为负，则立即返回函数。 如果您发现任何答案/代码不正确，或者您想分享更多关于上述主题的信息，请写评论。
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