for (int i=6; i>0; i++) { System.out.println (i); }
public class Test { public static void main (String[] argv) { int i = 5; int[] A = {6}; change (i); System.out.println (i); change (A[0]); System.out.println (A[0]); } static void change (int i) { i = i + 1; } }prints
for (int i=1; i<A.length; i++) { A[i] = A[i-1]; } System.out.println (Arrays.toString(A));results in:
public class Test { public static void main (String[] argv) { char[] letters = {'b','a','d','c','a','f','e'}; String s = strangeMethod (letters); System.out.println (s); } static String strangeMethod (char[] A) { int base = (int) 'a'; String s = ""; for (int i=0; i<A.length; i++) { int k = (int) A[i]; if (k < base+2) { s = s + A[i]; } } return s; } }This results in:
char[] letters = {'a','b','c','d','e','f'};write a method called mixHalves to return a string with all the odd-positioned letters together followed the even-positioned letters. Thus, for the above, with the call
String s = maxHalves(letters); System.out.println (s);you should get bdface as the printed string.
int[] A = {1, 2, 3, 2, 1}; int[] B = {1, 4, 9, 9, 4, 1}; int[] C = {1, 4, 9, 6, 1}; System.out.println (checkPalindrome(A)); System.out.println (checkPalindrome(B)); System.out.println (checkPalindrome(C));should print
true true falseWrite the method checkPalindrome using only while-loops (i.e., without for-loops).