blocks|key|3363139|text|我就是知道我会被否决的，但这里有一个解决方案。不是最好的，但你应该能够让它变得更好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3363140|class+Program
{
++++static+void+Main(string[]+args)
++++{
++++++++printNumDown(10);
++++}

++++static+void+printNumDown(int+num)
++++{
++++++++Console.WriteLine(num.ToString());
++++++++if+(num+>+1)
++++++++++++printNumDown(num+-+1);
++++++++else
++++++++++++printNumUp(num+%2B+1);
++++}

++++static+void+printNumUp(int+num)
++++{
++++++++Console.WriteLine(num.ToString());
++++++++if+(num+<+10)
++++++++++++printNumUp(num+%2B+1);
++++}
}|code-block|syntax|javascript|3363141|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I'm going to get downvoted I just know it but here is (a) solution. Not the best one but you should be able to make it better yourself.

<pre><code>class Program
{
 static void Main(string[] args)
 {
 printNumDown(10);
 }

 static void printNumDown(int num)
 {
 Console.WriteLine(num.ToString());
 if (num &gt; 1)
 printNumDown(num - 1);
 else
 printNumUp(num + 1);
 }

 static void printNumUp(int num)
 {
 Console.WriteLine(num.ToString());
 if (num &lt; 10)
 printNumUp(num + 1);
 }
}
</code></pre>

blocks|key|496773|text|void+print_recursive(int+n)+{+
++++printf("%25d\n",+n);
++++if+(n+<+10)
++++++++print_recursive(n%2B1);
++++printf("%25d\n",+n);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|496774|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>void print_recursive(int n) { 
 printf("%d\n", n);
 if (n &lt; 10)
 print_recursive(n+1);
 printf("%d\n", n);
}
</code></pre>

blocks|key|3363198|text|仅使用one+function和one+variable：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3363199|void+recurs(int+num)+{
++++printf("%25d\n",+num);
++++if+(num+<+10)+{
++++++++recurs(num+%2B+1);
++++}
++++printf("%25d\n",+num);
}

int+main()+{
++++recurs(1);
++++return+0;
}|code-block|syntax|javascript|3363200|entityMap^0|3|C|G|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

With one function and one variable only:

<pre><code>void recurs(int num) {
 printf("%d\n", num);
 if (num &lt; 10) {
 recurs(num + 1);
 }
 printf("%d\n", num);
}

int main() {
 recurs(1);
 return 0;
}
</code></pre>

blocks|key|495192|text|您应该能够自己弄清楚这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|495193|提示:创建一个接受10作为参数的方法，然后打印该参数并使用9调用自身。|offset|length|style|CODE|495194|entityMap^0|0|9|2|T|1|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

You should be able to figure this out yourself.

Hint: Make a method that takes a <code>10</code> as a parameter, then prints the parameter and calls itself with <code>9</code>.

blocks|key|1955161|text|这是一种偷偷摸摸的方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1955162|#include+<stdio.h>

static+void+recur_up+(int+n)+{
++++if+(n+>+1)
++++++++recur_up+(n+-+1);
++++printf+("%25d\n",+n);
}

static+void+recur_down+(int+n)+{
++++printf+("%25d\n",+n);
++++if+(n+>+1)
++++++++recur_down+(n+-+1);
}

int+main+(void)+{
++++recur_up+(10);
++++recur_down+(10);
++++return+0;
}|code-block|syntax|javascript|1955163|这将生成：|1955164|1
2
3
4
5
6
7
8
9
10
10
9
8
7
6
5
4
3
2
1|1955165|1955166|向下走比向上走要优雅得多，因为你可以用一个函数来做到这一点：|1955167|static+void+recur_both+(int+n)+{
++++printf+("%25d\n",+n);
++++if+(n+>+1)
++++++++recur_down+(n+-+1);
++++printf+("%25d\n",+n);
}|1955168|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|W|8|@]|9|@]|A|$]]|$1|N|3|O|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|P|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|Q|$]]

Here's a sneaky way:

<pre><code>#include &lt;stdio.h&gt;

static void recur_up (int n) {
 if (n &gt; 1)
 recur_up (n - 1);
 printf ("%d\n", n);
}

static void recur_down (int n) {
 printf ("%d\n", n);
 if (n &gt; 1)
 recur_down (n - 1);
}

int main (void) {
 recur_up (10);
 recur_down (10);
 return 0;
}
</code></pre>

which generates:

<pre><code>1
2
3
4
5
6
7
8
9
10
10
9
8
7
6
5
4
3
2
1
</code></pre>

<hr>

It would have been a lot more elegant going down then up since you could do that with a single function:

<pre><code>static void recur_both (int n) {
 printf ("%d\n", n);
 if (n &gt; 1)
 recur_down (n - 1);
 printf ("%d\n", n);
}
</code></pre>

blocks|key|1240416|text|Javascript版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1240417|printNumber(1);

function+printNumber(num){
++document.write(num);
++if+(num+<+10)+
+++++printNumber(num+%2B+1);
++document.write(num);
}|code-block|syntax|javascript|1240418|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Javascript version:

<pre><code>printNumber(1);

function printNumber(num){
 document.write(num);
 if (num &lt; 10) 
 printNumber(num + 1);
 document.write(num);
}
</code></pre>

blocks|key|1243580|text|为什么你们都这么难相处？在伪代码中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1243581|function+recurfunc(n)+{
++++if+(n+<+10)+{
++++echo+(-1+*+(floor(abs(n))+-+10));
++++++++recurfunc(n%2B1);
++++}
}|code-block|syntax|javascript|1243582|然后调用以-9.5为开头的递归函数。|1243583|在我看来，答案是使用绝对值，这似乎是显而易见的。|1243584|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Why are you guys all being so difficult? In Pseudocode:

<pre><code>function recurfunc(n) {
 if (n &lt; 10) {
 echo (-1 * (floor(abs(n)) - 10));
 recurfunc(n+1);
 }
}
</code></pre>

Then call recurfunc with -9.5 as its start.

Seems kind of obvious to me that the answer is using absolute value.

blocks|key|3363266|text|下面是Ruby中的一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3363267|puts+(r+=+->n=1+{+if+n<=10+then+[n]+%2B+r.(n%2B1)+%2B+[n]+else+[]+end+}).()|code-block|syntax|javascript|3363268|注意:在得出Ruby比Perl更难读的结论之前，让我向您保证:这不是惯用的Ruby。惯用的Ruby更像是|3363269|def+recursive_count_up_and_down(n=1)
++return+[]+unless+n<=10
++[n]+%2B
++recursive_count_up_and_down(n+%2B+1)+%2B
++[n]
end

puts+recursive_count_up_and_down|3363270|当然，Ruby是一种命令式语言，所以使用惯用法不会使用递归：|3363271|1.upto(10)+do+%7Ci%7C+puts+i+end
10.downto(1)+do+%7Ci%7C+puts+i+end|3363272|这是另一个很好的例子，不幸的是它也没有使用递归：|3363273|puts+Array.new(20)+{%7Ci%7C+if+i+<+10+then+i%2B1+else+20-i+end+}|3363274|顺便说一下:到目前为止，所有的解决方案，包括我的，实际上都是作弊的，因为从技术上讲，它们使用了两个变量，因为|3363275|function+foo+{}|3363276|等同于|3363277|var+foo+=+λ{}|3363278|因此，在上面的示例中，有两个变量：recursive_count_up_and_down和n。我们可以通过使用默认的无指针风格(例如，在SK演算中)来消除这两种情况，但我将把它留给读者作为练习。(意思是我想不出来:-)|offset|length|style|CODE|3363279|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|H|R|19|1|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1K|8|@$12|1L|13|1M|14|15]|$12|1N|13|1O|14|15]]|9|@]|A|$]]|$1|16|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|17|$]]

Here's one in Ruby:

<pre><code>puts (r = -&gt;n=1 { if n&lt;=10 then [n] + r.(n+1) + [n] else [] end }).()
</code></pre>

Note: before you conclude that Ruby is an unreadable mess even worse than Perl, let me assure you: this is not idiomatic Ruby. Idiomatic Ruby would be more like

<pre><code>def recursive_count_up_and_down(n=1)
 return [] unless n&lt;=10
 [n] +
 recursive_count_up_and_down(n + 1) +
 [n]
end

puts recursive_count_up_and_down
</code></pre>

Of course, Ruby is an imperative language, so doing it really idiomatically would not use recursion:

<pre><code>1.upto(10) do |i| puts i end
10.downto(1) do |i| puts i end
</code></pre>

Here's another neat one that unfortunately doesn't use recursion, either:

<pre><code>puts Array.new(20) {|i| if i &lt; 10 then i+1 else 20-i end }
</code></pre>

BTW: all the solutions so far, including mine, are actually cheating, because technically they use two variables, because

<pre><code>function foo {}
</code></pre>

is equivalent to

<pre><code>var foo = λ{}
</code></pre>

So, in my example above, there are two variables: <code>recursive_count_up_and_down</code> and <code>n</code>. We could eliminate both of those by writing in a tacit point-free style (for example in the SK-calculus), but I'll leave that as an exercise to the reader. (Meaning I can't figure it out :-) )

blocks|key|1243586|text|#include<stdio.h>
void+c(int+n)
{
+++static+int+i;
+++if(i<n)
+++{
+++++i%2B%2B;
+++++printf("%25d",i);
+++++c(n);
+++}
}
int+main()
{
++c(5);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1243587|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include&lt;stdio.h&gt;
void c(int n)
{
 static int i;
 if(i&lt;n)
 {
 i++;
 printf("%d",i);
 c(n);
 }
}
int main()
{
 c(5);
}
</code></pre>

blocks|key|1955321|text|导入java.util.Scanner；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1955322|公共类PrintNumberFrom1To10{|1955323|public+static+void+printNumber(int+num){


++++if(num<=10){


++++System.out.println(num);


++++printNumber(num%2B1);


++++}
}

public+static+void+main(String[]+args)+{


++++System.out.println("Enter+Your+Number:+");


++++Scanner+sc+=+new+Scanner(System.in);



++++int+num+=+sc.nextInt();




++++printNumber(num);






}|code-block|syntax|javascript|1955324|}|1955325|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

import java.util.Scanner;

public class PrintNumberFrom1To10{

<pre><code>public static void printNumber(int num){


 if(num&lt;=10){


 System.out.println(num);


 printNumber(num+1);


 }
}

public static void main(String[] args) {


 System.out.println("Enter Your Number: ");


 Scanner sc = new Scanner(System.in);



 int num = sc.nextInt();




 printNumber(num);






}
</code></pre>

}

I am preparing for my interview tomorrow -- I need the answer to this question:

How can you print 1 to 10 &amp; 10 to 1 by using recursion with single variable

How can you print 1 to 10 & 10 to 1 by using recursion with single variable

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在准备明天的面试--我需要这个问题的答案：如何使用带有单变量的递归打印1到10和10到1

问如何使用带有单变量的递归打印1到10和10到1
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回答 10

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问如何使用带有单变量的递归打印1到10和10到1EN

回答 10

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问如何使用带有单变量的递归打印1到10和10到1
EN