Reputation: 11
public class Main {
public static void main(String[] args) {
print(10);
}
static void print(int s)
{
if (1==s) {
System.out.print(s);
}
System.out.print(s);
print(s-1);
}
}
But I want output like this:
12345678910
Upvotes: 1
Views: 188
Reputation: 929
Solution 1:- If you want the end result-oriented then you can try this
public static int doCalculaton(int numVal, int endVal) {
System.out.println(endVal - numVal);
if (numVal == 0) {
return numVal;
}
return doCalculaton(numVal - 1, endVal);
}
public static void main(String[] args) {
int toBeCalculateNum = 10;
doCalculaton(toBeCalculateNum, toBeCalculateNum);
}
Solution 2:- If you want to use the result after calling
public static int doCalculaton(int numVal, ArrayList<Integer> numerList) {
numerList.add(numVal);
if (numVal == 0) {
return numVal;
}
return doCalculaton(numVal - 1, numerList);
}
public static void main(String[] args) {
ArrayList<Integer> numerList = new ArrayList<>();
doCalculaton(10, numerList);
Collections.reverse(numerList);
System.out.println(numerList);
// TODO loop and do whatever you want to do
}
Upvotes: 0
Reputation: 105
Here, try this.
public static void main(String[] args) {
print(1, 10);
}
static void print(int startValue, int endValue)
{
System.out.print(startValue);
if(startValue < endValue)
print(startValue+1, endValue);
}
Upvotes: 1
Reputation: 393801
You existing recursion never ends. You should only make the recursive call if s >= 1
.
And to print the numbers in increasing order, you need to first make the recursive call and then print the current number:
static void print(int s)
{
if (s >= 1) {
print(s-1);
System.out.print(s);
}
}
Upvotes: 2