Question
Write a method called writeSequence that accepts an integer n as a parameter and prints to the console a symmetric sequence of n numbers composed of descending integers that ends in 1, followed by a sequence of ascending integers that begins with 1. The following table indicates the output that should be produced for various values of n:
Method call Output produced
-----------------------------------------
writeSequence(1); 1
writeSequence(2); 1 1
writeSequence(3); 2 1 2
writeSequence(4); 2 1 1 2
writeSequence(5); 3 2 1 2 3
writeSequence(6); 3 2 1 1 2 3
writeSequence(7); 4 3 2 1 2 3 4
writeSequence(8); 4 3 2 1 1 2 3 4
writeSequence(9); 5 4 3 2 1 2 3 4 5
writeSequence(10); 5 4 3 2 1 1 2 3 4 5
Notice that when n is odd the sequence has a single 1 in the middle, while for even values it has two 1s in the middle. Your method should throw an IllegalArgumentException if it is passed a value less than 1.
Answer
package recursion;
public class WriteSequence {
public static void helper1(int value) {
System.out.print((value / 2) + " ");
writeSequence(value - 2);
System.out.print(" " + (value / 2));
}
public static void helper2(int value) {
System.out.print(((value + 1) / 2) + " ");
writeSequence((value - 2));
System.out.print(" " + ((value + 1) / 2));
}
public static void writeSequence(int value) {
if (value < 1)
throw new IllegalArgumentException();
if (value == 1)
System.out.print("1");
else if(value == 2) {
System.out.print("1 1");
}
// we use helper function to write the sequence based on even or odd number
else if(value%2==0) {
helper1(value);
}else
helper2(value);
}
public static void main(String[] args) {
int x = 10;
System.out.print("writeSequence("+x+"): ");
writeSequence(x);
}
}
Output:
writeSequence(10): 5 4 3 2 1 1 2 3 4 5