递归练习题|第 2 集
原文:https://www . geesforgeks . org/practice-questions-for-recursion-set-2/
解释以下功能的功能。
问题 1
C
/* Assume that n is greater than or equal to 1 */
int fun1(int n)
{
if (n == 1)
return 0;
else
return 1 + fun1(n / 2);
}
Java 语言(一种计算机语言，尤用于创建网站)
/* Assume that n is greater than or equal to 1 */
static int fun1(int n)
{
if (n == 1)
return 0;
else
return 1 + fun1(n / 2);
}
// This code is contributed by shubhamsingh10
Python 3
# Assume that n is greater than or equal to 1 */
def fun1(n):
if(n == 1):
return 0
else:
return 1 + fun1(n//2)
# This code is contributed by shubhamsingh10
C
/* Assume that n is greater than or equal to 1 */
static int fun1(int n)
{
if (n == 1)
return 0;
else
return 1 + fun1(n / 2);
}
// This code is contributed by shubhamsingh10
java 描述语言
<script>
/* Assume that n is greater than or equal to 1 */
function fun1(n)
{
if (n == 1)
return 0
else
return 1 + fun1(n / 2)
}
// This code is contributed by gottumukkalabobby
</script>
C++
/* Assume that n is greater than or equal to 1 */
int fun1(int n)
{
if (n == 1)
return 0;
else
return 1 + fun1(n / 2);
}
// This code is contributed by shubhamsingh10
// Improved by Adwitiya Mourya
回答:函数计算并返回
。例如，如果 n 在 8 和 15 之间，那么 fun1()返回 3。如果 n 在 16 到 31 之间，则 fun1()返回 4。 问题 2
C++
/* Assume that n is greater than or equal to 0 */
void fun2(int n)
{
if(n == 0)
return;
fun2(n/2);
cout << n%2 << endl;
}
//This code is contributed by shubhamsingh10
C
/* Assume that n is greater than or equal to 0 */
void fun2(int n)
{
if(n == 0)
return;
fun2(n/2);
printf("%d", n%2);
}
Java 语言(一种计算机语言，尤用于创建网站)
/* Assume that n is greater than or equal to 1 */
static void fun2(int n)
{
if(n == 0)
return;
fun2(n/2);
System.out.println(n%2);
}
// This code is contributed by Shubhamsingh10
Python 3
# Assume that n is greater than or equal to 0 */
def fun2(n):
if(n == 0):
return
fun2(n / 2)
print(n % 2, end="")
# This code is contributed by shubhamsingh10
C
void fun2(int n)
{
if(n == 0)
return;
fun2(n/2);
Console.Write(n%2);
}
// This code is contributed by shubhamsingh10
java 描述语言
<script>
// Assume that n is greater than or equal to 1
function fun2(n)
{
if (n == 0)
return;
fun2(n / 2);
document.write(n % 2 + " ")
}
// This code is contributed by gottumukkalabobby
</script>
回答:fun2()函数打印 n 的二进制等价物。例如，如果 n 是 21，那么 fun2()打印 10101。 注意，上面的函数只是用来练习递归的，它们并不是它们提供的功能的理想实现。 如发现任何答案/代码不正确，请写评论。
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