Prentice.Hall.Introduction.to.Java.Programming,.Brief.Version.9th.(2014).[sharethefiles.com]

20.4 Problem Solving Using Recursion 745
Recursion is everywhere. It is fun to think recursively. Consider drinking coffee. You may
describe the procedure recursively as follows:
think recursively
public static void drinkCoffee(Cup cup) {
if (!cup.isEmpty()) {
cup.takeOneSip(); // Take one sip
drinkCoffee(cup);
}
}
Assume cup is an object for a cup of coffee with the instance methods isEmpty() and
takeOneSip(). You can break the problem into two subproblems: one is to drink one sip of
coffee and the other is to drink the rest of the coffee in the cup. The second problem is the
same as the original problem but smaller in size. The base case for the problem is when the
cup is empty.
Consider the problem of printing a message n times. You can break the problem into two
subproblems: one is to print the message one time and the other is to print it n - 1 times. The
second problem is the same as the original problem but it is smaller in size. The base case for
the problem is n == 0. You can solve this problem using recursion as follows:
public static void nPrintln(String message, int times) {
if (times >= 1) {
System.out.println(message);
nPrintln(message, times - 1);
} // The base case is times == 0
}
Note that the fib method in the preceding section returns a value to its caller, but the
drinkCoffee and nPrintln methods are void and they do not return a value.
If you think recursively, you can use recursion to solve many of the problems presented in
earlier chapters of this book. Consider the palindrome problem in Listing 9.1. Recall that a
string is a palindrome if it reads the same from the left and from the right. For example,
“mom” and “dad” are palindromes, but “uncle” and “aunt” are not. The problem of checking
whether a string is a palindrome can be divided into two subproblems:
■ Check whether the first character and the last character of the string are equal.
■ Ignore the two end characters and check whether the rest of the substring is a
palindrome.
The second subproblem is the same as the original problem but smaller in size. There are two
base cases: (1) the two end characters are not the same, and (2) the string size is 0 or 1. In case
1, the string is not a palindrome; in case 2, the string is a palindrome. The recursive method
for this problem can be implemented as shown in Listing 20.3.
recursive call
think recursively
LISTING 20.3
RecursivePalindromeUsingSubstring.java
1 public class RecursivePalindromeUsingSubstring {
2
3
public static boolean isPalindrome(String s) {
if (s.length()