blocks|key|4625704|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4625705|public+static+int+howManyChar+(String+s,+char+c)+{+++++if+(s.length()==0)+++++++++return+0;+++++return+s.charAt(0)+==+c+?+1+:+0+%2B+howManyChar(s.substring(1),+c);+}|offset|length|style|CODE|4625706|或者如果您不能使用substring方法|4625707|public+static+int+howManyChar+(String+s,+char+c,+int+pos)+{+++++if+(pos+>=+s.length())+++++++++return+0;+++++return+s.charAt(pos)+==+c+?+1+:+0+%2B+howManyChar(s,+c,+%2B%2Bpos);+}|4625708|entityMap^0|0|0|4J|0|0|0|4S|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|M|$]]

try this

<code>public static int howManyChar (String s, char c) {
 if (s.length()==0)
 return 0;
 return s.charAt(0) == c ? 1 : 0 + howManyChar(s.substring(1), c);
}</code>

OR if you cannot use the substring method

<code>public static int howManyChar (String s, char c, int pos) {
 if (pos &gt;= s.length())
 return 0;
 return s.charAt(pos) == c ? 1 : 0 + howManyChar(s, c, ++pos);
}</code>

blocks|key|119851|text|public+static+int+howManyChar+(String+s,+char+c)
{
++++index%2B%2B;
++++++++if+(s.length()+==+index)
++++++++++++return+0;
++++++++else+if(s.charAt(index)+==+c)
+++++++++++return+1+%2B+howManyChar(s,+c);
++++++++else
++++++++++++return+0+%2B++howManyChar(s,+c);

++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|119852|现在这不是最好的解决方案，我有一个类级别的变量(index)来保存字符串中的位置。但是如果不使用子字符串，我想不出更好的方法。|unstyled|119853|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>public static int howManyChar (String s, char c)
{
 index++;
 if (s.length() == index)
 return 0;
 else if(s.charAt(index) == c)
 return 1 + howManyChar(s, c);
 else
 return 0 + howManyChar(s, c);

 }
</code></pre>

Now this isn't the best soultion, I have a class level variable (index) holding the place in the string. But without using substring I can't think of a better way to do it.

blocks|key|4625766|text|我有一个没有下标的很好的解决方案...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4625767|public+class+test{

+++int+counter=0;int+last;
++++public+static+void+main(String[]+args){
++++++++test+t1=new+test();+//object+of+test
++++++++String+line="This+is+a+working+code";+//line+in+which+char+to+be+counted
++++++++char+c='q';+//+char+to+be+counted
++++++++t1.last=line.length()-1;++//setting+last
++++++++System.out.println("In+sentence:+"%2Bline);
++++++++if(t1.count(line,+c)!=-1)
++++++++System.out.println("The+character+"%2Bc%2B"+is+"%2Bt1.counter%2B"+times");+//printing+result
++++++++else
++++++++System.out.println("The+character+"%2Bc%2B"+is+not+found");
++++}
++++public+int+count+(String+line,+char+c){
++++++++if+(last<0){+++//+case+for+empty+string+or+nil+char.
++++++++++++return+counter;+++++++++++++++++++++//+recursion+ends+here+and+return+counted
++++++++}
++++++++else+if(line.charAt(last)==c)
++++++++{
++++++++++++counter%2B%2B;++//counting
++++++++}
++++++++last--;++//shitin+last;
++++++++count(line,+c);
++++++++return+-1;++//if+even+a+single+times+is+not+found+then+return+-1;
++++++}
}|code-block|syntax|javascript|4625768|它可以帮助您从命令prompt......Thank+you...You+...获取要搜索的行和字符。|4625769|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|$]]

I have a good solution without subscript...

<pre><code>public class test{

 int counter=0;int last;
 public static void main(String[] args){
 test t1=new test(); //object of test
 String line="This is a working code"; //line in which char to be counted
 char c='q'; // char to be counted
 t1.last=line.length()-1; //setting last
 System.out.println("In sentence: "+line);
 if(t1.count(line, c)!=-1)
 System.out.println("The character "+c+" is "+t1.counter+" times"); //printing result
 else
 System.out.println("The character "+c+" is not found");
 }
 public int count (String line, char c){
 if (last&lt;0){ // case for empty string or nil char.
 return counter; // recursion ends here and return counted
 }
 else if(line.charAt(last)==c)
 {
 counter++; //counting
 }
 last--; //shitin last;
 count(line, c);
 return -1; //if even a single times is not found then return -1;
 }
}
</code></pre>

It may help you...You may take line and char which is to be searched as input from Command prompt......Thank you....

blocks|key|119947|text|我已经看过这里的一些答案，它们要么对于您的问题来说太复杂，要么使用子字符串，要么不使用基本的递归|type|unstyled|depth|inlineStyleRanges|entityRanges|data|119948|您的解决方案应该是这样的。|119949|function+countChar(String+a,+char+b,+int+i)
if(i==a.length)
return+0
else+if(b+==a.charAt(i))
return+1+%2B+function+countChar(a,+b,+i%2B1)
else
return+function+countChar(a,b,i%2B1)|code-block|syntax|javascript|119950|这部分代码就是循环。|119951|function+countChar(String+a,+char+b,+int+i)
{
+++if(i==a.length)+
+++{
+++++++return+0
+++}
+++else+
+++{
++++++//put+here+a+print+statement+to+see+what+happens.
+++++//+print+charAt(i)+or+i+itself
++++++return+function+countChar(a,+b,+i%2B1)+//<--+moving+to+the+next+index
+++}
}|119952|学习这个循环。您应该使用charAt(i)。子字符串不是必需的..|119953|为了满足需求，您可以将您的方法包装在另一个递归方法中。|119954|function+howManyChar(String+a,+Char+b)
{
+++countChar(a,+b,+0);
}|119955|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

I have looked at some of the answers here, they are either too complicated for your question, use substring, or do not use basic recursion

your solution should be something like this. 

<pre><code>function countChar(String a, char b, int i)
if(i==a.length)
return 0
else if(b ==a.charAt(i))
return 1 + function countChar(a, b, i+1)
else
return function countChar(a,b,i+1)
</code></pre>

This part of the code is the loop.

<pre><code>function countChar(String a, char b, int i)
{
 if(i==a.length) 
 {
 return 0
 }
 else 
 {
 //put here a print statement to see what happens.
 // print charAt(i) or i itself
 return function countChar(a, b, i+1) //&lt;-- moving to the next index
 }
}
</code></pre>

Learn that loop. You should use charAt(i). Substring is not neccesssary..

In order to fulfil the requirement, you can wrap your method in another recursive one. 

<pre><code>function howManyChar(String a, Char b)
{
 countChar(a, b, 0);
}
</code></pre>

i need a litle bit help.
i need to write this method recursively:
public static int howManyChar (String s, char c)
and without using the substring method and loops..

any ideas?
i am stuck at this point.

<pre><code>public static int howManyChar (String s, char c)
{
 int count = 0;
 int lastChar = s.length()-1;
 if (s.length()== 0)
 return 0;
 if (s.charAt(lastChar) == c)
 {
 count++;
 }
</code></pre>

i have succeeded using the substring method:

<pre><code> int count =0;
 if (s.charAt(0)==c)
 count++;
 if (s.length()&gt;1)
 {
 count+=howManyChar(s.substring(1), c);
 }
 return count;
</code></pre>

but i want to write the code without using substring.

Recursively counting chars in java

我需要一点帮助。我需要递归地编写此方法: public static int howManyChar (String s，char c)，并且不使用子字符串方法和循环。有什么想法吗？我被困在这一点上了。public static int howManyChar (String s, char c){    int co...

问在java中递归计算字符
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