I am very new to programming and I have to write a method and program for the following; public static String repeat(String str, int n) that returns the string repeated n times. Example ("ho", 3) returns "hohoho" Here is my program so far:
public static void main(String[] args) {
// **METHOD** //
Scanner in = new Scanner(System.in);
System.out.println("Enter a string");
String str = in.nextLine();
System.out.println(repeat (str));//Having trouble with this line
}
// **TEST PROGRAM**//
public static String repeat(String str, int n)
{
if (n <= 0)
{
return ""//Having trouble with this line
}
else if (n % 2 == 0)
{
return repeat(str+str, n/2);
}
else
{
return str + repeat(str+str, n/2);
}
}
}
I made some changes to my code, but it still is not working
public static void main(String[] args) {
// **METHOD** //
Scanner in = new Scanner(System.in);
System.out.println("Enter a string");
String str = in.nextLine();
int n = in.nextInt();
System.out.println(repeat(str,n));
}
// **TEST PROGRAM**//
public static String repeat(String str, int n)
{
if (n <= 0)
{
return "";
}
else if (n % 2 == 0)
{
return repeat(str+str, n/2);
}
else
{
return str + repeat(str+str, n/2);
}
}
}
You've missed a semi colon on the line you're having trouble with, it should be return "";
and not return ""
Also, the line System.out.println(repeat (str));
should have 2 arguments because you're repeat definition is:
public static String repeat(String str, int n)
As a further note, an easier function might be
public static String repeat(String str, int n) { if (n == 0) return ""; String return_str = ""; for (int i = 0; i < n; i++) { return_str += str; } return return_str; }
我很快注意到的两件事:您忘记了一个半列,而您对“ repeat”的调用与方法签名不匹配(您忘记了n)
You are not passing correct arguments while you calling the desired method.
Call as repeat (str,k)
k
- should be an integer
public static String repeat(String toRepeat, int n){
if(n==0){
return "";
}
return toRepeat+repeat(toRepeat,n-1);
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.