blocks|key|2305184|text|public+int+count7(int+n)+{
++int+counter+=+0;

++if(+n+%25+10+==+7)+counter%2B%2B;

++if(+n+/+10++==+0)++return+counter;

++return+counter+%2B+count7(n/10);+
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2305185|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>public int count7(int n) {
 int counter = 0;

 if( n % 10 == 7) counter++;

 if( n / 10 == 0) return counter;

 return counter + count7(n/10); 
}
</code></pre>

blocks|key|2305133|text|好的，这是我写的解决方案，让我们看看如何只返回一次|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2305134|public+int+count7(int+n)+
{
++++int+c+=+0;
++++if+(7+>+n)
++++{
++++++++return+0;
++++}
++++else
++++{
++++++++if+(+7+==+n+%25+10)
++++++++{
++++++++++++c+=+1;
++++++++}
++++++++else
++++++++{
++++++++++++c+=+0;
++++++++}
++++}
++++return+c+%2B+count7(n+/+10);++
}|code-block|syntax|javascript|2305135|同样，只有一次返回|2305136|public+int+count7(int+n)+
{
++++return+(7+>+n)+?+0+:+(+(+7+==+n+%25+10)+?+1+%2B+count7(n+/+10)+:+0+%2B+count7(n+/+10));++
}|2305137|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/spookiecookie/codingbatsolutions/blob/master/codingbat/recursion1/Count7.java|1|http://ideone.com/lHBxec^0|5|7|0|0|0|0|9|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|K|3|L|5|F|7|15|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

Well here's <a href="https://github.com/spookiecookie/codingbatsolutions/blob/master/codingbat/recursion1/Count7.java" rel="noreferrer">the solution I wrote</a> and let's see how to do it with only one return

<pre><code>public int count7(int n) 
{
 int c = 0;
 if (7 &gt; n)
 {
 return 0;
 }
 else
 {
 if ( 7 == n % 10)
 {
 c = 1;
 }
 else
 {
 c = 0;
 }
 }
 return c + count7(n / 10); 
}
</code></pre>

<a href="http://ideone.com/lHBxec" rel="noreferrer">the same with only one return</a>

<pre><code>public int count7(int n) 
{
 return (7 &gt; n) ? 0 : ( ( 7 == n % 10) ? 1 + count7(n / 10) : 0 + count7(n / 10)); 
}
</code></pre>

blocks|key|2026473|text|当然，PFB我的解决方案在JAVA中也是如此|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2026474|public+int+count7(int+n)+{
+++if((n+/+10+==+0)+&&+!(n+%25+10+==+7))++++++//First+BASE+CASE+when+the+left+most+digit+is+7+return+1
+++++++return+0;+++++
+++else+if((n+/+10+==+0)+&&+(n+%25+10+==+7))+//Second+BASE+CASE+when+the+left+most+digit+is+7+return+0
++++++++return+1;
+++else+if((n+%25+10+==+7))+++
+++//if+the+number+having+2+digits+then+test+the+rightmost+digit+and+trigger+recursion+trimming+it+there++++++
+++++++return+1+%2B+count7(n+/+10);
+++return+count7(n+/+10);
}|code-block|syntax|javascript|2026475|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|$]]

Sure PFB my solution in JAVA for the same

<pre><code>public int count7(int n) {
 if((n / 10 == 0) &amp;&amp; !(n % 10 == 7)) //First BASE CASE when the left most digit is 7 return 1
 return 0; 
 else if((n / 10 == 0) &amp;&amp; (n % 10 == 7)) //Second BASE CASE when the left most digit is 7 return 0
 return 1;
 else if((n % 10 == 7)) 
 //if the number having 2 digits then test the rightmost digit and trigger recursion trimming it there 
 return 1 + count7(n / 10);
 return count7(n / 10);
}
</code></pre>

blocks|key|1307194|text|public+int+count7(int+n)+{+

++if(n+==+0)
++++++return+0;
++else{
+++++if(n%2510+==7)
+++++++++return+1%2Bcount7(n/10);

+++++return+0%2Bcount7(n/10);++
++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1307195|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>public int count7(int n) { 

 if(n == 0)
 return 0;
 else{
 if(n%10 ==7)
 return 1+count7(n/10);

 return 0+count7(n/10); 
 }
}
</code></pre>

blocks|key|1307214|text|public+static+int+count7(int+n){
++++++++if(n+==+7)
++++++++++++return+1;
++++++++else+if(n+>+9){
++++++++++++int+a+=+count7(n%2510);
++++++++++++int+b+=+count7(n/10);
++++++++++++return+a+%2B+b;
++++++++}else
++++++++++++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1307215|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>public static int count7(int n){
 if(n == 7)
 return 1;
 else if(n &gt; 9){
 int a = count7(n%10);
 int b = count7(n/10);
 return a + b;
 }else
 return 0;
}
</code></pre>

blocks|key|2305143|text|public+int+count7(int+n)+
{++
+if(n==0)return+0;
+if(n%2510==7)return+1%2Bcount7(n/10);
+else+return+count7(n/10);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2305144|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>public int count7(int n) 
{ 
 if(n==0)return 0;
 if(n%10==7)return 1+count7(n/10);
 else return count7(n/10);
}
</code></pre>

blocks|key|2305174|text|public+int+count7(int+n)+{
++if+(n+!=+7+&&+n+<+10)+return+0;
++else+if+(n+==+7)+return+1;
++else+if+(n%2510+==+7)+return+count7(n/10)+%2B+1+;
++else+return+count7(n/10);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2305175|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>public int count7(int n) {
 if (n != 7 &amp;&amp; n &lt; 10) return 0;
 else if (n == 7) return 1;
 else if (n%10 == 7) return count7(n/10) + 1 ;
 else return count7(n/10);
}
</code></pre>

blocks|key|2026520|text|public+int+count7(int+n){
++++if(n+<+7)
++++++++return+0;
++++else+if(n+%25+10+==+7)
++++++++return+1+%2B+count7(n+/+10);
++++else
++++++++return+count7(n+/+10);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2026521|第一个if-语句是我们希望终止的基本情况。第二次检查看最右边的数字是否是7，如果是，把最右边的数字砍下来，然后再试一次。当递归调用终止并开始在链上返回值时，添加1以包含此成功的检查。如果上述语句都不是真语句，请剪掉最右边的数字，然后再试一次。|unstyled|2026522|我知道这是2年前，但希望这是一个更易读和直观，因此有帮助。|2026523|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>public int count7(int n){
 if(n &lt; 7)
 return 0;
 else if(n % 10 == 7)
 return 1 + count7(n / 10);
 else
 return count7(n / 10);
}
</code></pre>

the first if-statement is the base case that we would want to terminate on. The second checks to see if the rightmost digit is 7. If it is, chop the rightmost digit off and try again. When the recursive calls terminate and and values start getting returned up the chain, add 1 to include this successful check. In the case that neither of the above statements are true, chop off the rightmost digit and try again.

I know this is 2 years old, but hope this is a bit more readable and intuitive, and thus helpful.

blocks|key|2026719|text|我就是这么做的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2026720|public+int+count7(int+n)+{
++return+(n==0?0:(count7(n/10)%2B(n%2510==7?1:0)));
}|code-block|syntax|javascript|2026721|递归转到0，返回0，当它返回时，检查每个数字是否为7，如果它是0，则返回1。它继续增加这些，直到所有的出路。|2026722|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here is how I did it.
<pre><code>public int count7(int n) {
 return (n==0?0:(count7(n/10)+(n%10==7?1:0)));
}
</code></pre>
Recursion goes to 0 and it returns 0, when it comes back out checks if each number is 7, returns 1 if it is else 0. It continues adding those until all the way out.

blocks|key|2305223|text|使用一次返回可能会增加阅读难度。如果要计算递归中出现的次数，一个简单的公式是创建一个要终止的基大小写，然后提供一个增量返回，最后是一个返回，该返回将有助于在不递增的情况下到达基本情况。例如..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2305224|public+int+count7(int+n)+{
++if(n+==+0)+return+0;
++if(n+%25+10+==+7)+return+1+%2B+count7(n+/+10);
++return+count7(n+/+10);
}|code-block|syntax|javascript|2305225|在我看来，使用像下面这样的一行返回很难读懂或更新，因为双三值。|2305226|public+int+count7(int+n)+
{
++++return+(n+==+0)+?+0+:+(n+%25+10+==+7)+?+1+%2B+count7(n+/+10)+:+count7(n+/+10);++
}|2305227|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Using one return will likely make it harder to read. If you are counting occurrences in recursion, an easy formula is to create a base case to terminate on, then provide an incremental return, and finally a return that will aid in reaching the base case without incrementing. For example..

<pre><code>public int count7(int n) {
 if(n == 0) return 0;
 if(n % 10 == 7) return 1 + count7(n / 10);
 return count7(n / 10);
}
</code></pre>

Using a one liner return like the one below in my opinion is harder to read or update because of the double ternary..

<pre><code>public int count7(int n) 
{
 return (n == 0) ? 0 : (n % 10 == 7) ? 1 + count7(n / 10) : count7(n / 10); 
}
</code></pre>

blocks|key|2305285|text|通过输入的模数，我的解决方案从第n位向第一位倒转。我们将返回中找到的七点数加到最后的输出中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2305286|然后检查输入是否小于7可以是下一步。如果输入小于7，那么从一开始输入中就没有7s。|2305287|public+int+count7(int+n)+{
++++++++int+sevens_found+=+0;
++++++++if(+n+%25+10+==+7+)+sevens_found+%2B%2B;
++++++++++++return+(+n+<+7)+?+0+:+(+n+%25+10+==+7+)+?+sevens_found+%2B+count7+(+n+/+10+)+:+count7+(+n+/+10+);
++++++++}|code-block|syntax|javascript|2305288|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

My solution works backwards from the nth digit to the first digit, by taking the modulus of the input. we add the number of sevens found into the return, for the final output. 

Then checking whether the input is less than 7 can be the next step. If the input is less than 7 then there have never been any 7s in the input to begin with.

<pre><code>public int count7(int n) {
 int sevens_found = 0;
 if( n % 10 == 7 ) sevens_found ++;
 return ( n &lt; 7) ? 0 : ( n % 10 == 7 ) ? sevens_found + count7 ( n / 10 ) : count7 ( n / 10 );
 }
</code></pre>

blocks|key|1307283|text|#include<bits/stdc%2B%2B.h>
using+namespace+std;

int+count_occurences(int+k){
++++int+r;
++++if(k==0){
++++++++return+0;
++++}
++++r+=+k%2510;+++
++++k+=+k/10;

++++if(r!=7){
++++++++return+count_occurences(k);
++++}
++++return+1%2Bcount_occurences(k);
++++}



int+main()
{
++++int+x;
++++cin>>x;
++++++cout<<"+"<<count_occurences(x);

return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1307284|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;bits/stdc++.h&gt;
using namespace std;

int count_occurences(int k){
 int r;
 if(k==0){
 return 0;
 }
 r = k%10; 
 k = k/10;

 if(r!=7){
 return count_occurences(k);
 }
 return 1+count_occurences(k);
 }



int main()
{
 int x;
 cin&gt;&gt;x;
 cout&lt;&lt;" "&lt;&lt;count_occurences(x);

return 0;
}
</code></pre>

blocks|key|1307291|text|public+int+count7(int+n)+{
++++int+length+=+0;
++++int+counter+=+0;
++++if+((n+/+10)+*+10+!=+n+%7C%7C+(n+/+10)+!=+0)+{
++++++++if+(n+%25+10+!=+7)+{
++++++++++++counter%2B%2B;
++++++++}
++++++++length+%2B=+1+%2B+count7(n+/+10);
++++}
++++return+length+-+counter;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1307292|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>public int count7(int n) {
 int length = 0;
 int counter = 0;
 if ((n / 10) * 10 != n || (n / 10) != 0) {
 if (n % 10 != 7) {
 counter++;
 }
 length += 1 + count7(n / 10);
 }
 return length - counter;
}
</code></pre>

blocks|key|1307354|text|N+==+0的基本情况只是“使我们脱离了递归循环”，n%25+10+==+7允许我们实际计算整数中7s的数量，返回语句遍历给定的参数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1307355|public+int+count7(int+n)+{
+++if+(n+==+0)+
++++++return+0;
+++if+(n+%25+10+==+7)+
++++++return+1+%2B+count7(n+/+10);
+++return+count7(n+/+10);+
}|code-block|syntax|javascript|1307356|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|$]]

The base case of n == 0 just &quot;breaks&quot; us out of the recursive loop, the n % 10 == 7 allows us to actually count the amount of 7s in an integer, and the return statement iterates through the given argument.
<pre><code>public int count7(int n) {
 if (n == 0) 
 return 0;
 if (n % 10 == 7) 
 return 1 + count7(n / 10);
 return count7(n / 10); 
}
</code></pre>

blocks|key|2305322|text|public+int+count7(int+n)
{
++++return+occurrencesCounting(n,+0);
}

private+int+occurrencesCounting(int+n,+int+count)+
{
++++int+counter+=+n+%25+10+==+7+?+count+%2B+1+:+count;
++++if+(n+/+10+==+0)
++++{
++++++++return+counter;
++++}+++++
++++return+occurrencesCounting(n+/+10,+counter+);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2305323|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>public int count7(int n)
{
 return occurrencesCounting(n, 0);
}

private int occurrencesCounting(int n, int count) 
{
 int counter = n % 10 == 7 ? count + 1 : count;
 if (n / 10 == 0)
 {
 return counter;
 } 
 return occurrencesCounting(n / 10, counter );
}
</code></pre>

Can anybody help me programming the next problem (taken from <a href="http://codingbat.com/java" rel="nofollow">Codingbat</a>- <a href="http://codingbat.com/java/Recursion-1" rel="nofollow">Recursion1</a>- <a href="http://codingbat.com/prob/p101409" rel="nofollow">count7</a>)

Given a non-negative <code>int n,</code> return the count of the <code>occurrences of 7 as a digit</code>, so for example <code>717 yields 2.</code> (no loops). Note that <code>mod (%)</code> by 10 yields the <code>rightmost digit (126 % 10 is 6),</code> while divide (/) by 10 removes the rightmost digit (126 / 10 is 12). 

<pre><code>count7(717) → 2
count7(7) → 1
count7(123) → 0
</code></pre>

There are some solutions which includes number of "returns".
I would like to program the problem with only 1 "return".

Codingbat- Recursion1- count7

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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有人能帮我编写下一个问题(摘自- - )吗？给定一个非负的int n,，返回occurrences of 7 as a digit的计数，例如717 yields 2. (无循环)。注意，mod (%)乘以10生成rightmost digit (126 % 10 is 6),，而除法(/)减去最右边的数字(126 / 10是12)。count7(717) → 2count7(7) → 1coun

问Codingbat- Recursion1- count7
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回答 15

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