Question

this question might seem basic to some people but I’ve been analysing and dissecting this code without success as to how this permutation program by Robert Sedgewick prints the combination o f words or characters without using system.out.print in the methods perm1 and perm2. Any help or explanation for dummies is greatly appreciated. Thank you in advance.

This is the code under the link:

public class Permutations {
  // print N! permutation of the characters of the string s (in order)
  public  static void perm1(String s) { perm1("", s); }
  private static void perm1(String prefix, String s) {
      int N = s.length();
      if (N == 0) System.out.println(prefix);
      else {
          for (int i = 0; i < N; i++)
             perm1(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, N));
      }
  }
  // print N! permutation of the elements of array a (not in order)
  public static void perm2(String s) {
     int N = s.length();
     char[] a = new char[N];
     for (int i = 0; i < N; i++)
         a[i] = s.charAt(i);
     perm2(a, N);
  }

  private static void perm2(char[] a, int n) {
      if (n == 1) {
          System.out.println(a);
          return;
      }
      for (int i = 0; i < n; i++) {
          swap(a, i, n-1);
          perm2(a, n-1);
          swap(a, i, n-1);
      }
  }  

  // swap the characters at indices i and j
  private static void swap(char[] a, int i, int j) {
      char c;
      c = a[i]; a[i] = a[j]; a[j] = c;
  }

  public static void main(String[] args) {
     int N = Integer.parseInt(args[0]);
     String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
     String elements = alphabet.substring(0, N);
     perm1(elements);
     System.out.println();
     perm2(elements);
  }
}

Answer 1

It’s right there:

if (N == 0) System.out.println(prefix);

System.out.print and System.out.println is basically the same, except the later prints a newline after the text.