Write a recursive method called isReverse that accepts two strings as parameters and returns true if the two strings contain the same sequence of characters as each other but in the opposite order (ignoring capitalization), and false otherwise. For example, the call of isReverse("hello", "eLLoH") would return true. The empty string, as well as any one-letter string, is considered to be its own reverse

public static boolean isReverse(String s1, String s2)

{

if (s1.length() == 0 && s2.length() == 0)

{

return true;

}

else if (s1.length() == 0 || s2.length() == 0)

{

return false;

} else

{

String s1first = s1.substring(0, 1);

String s2last = s2.substring(s2.length() - 1);

return s1first.equalsIgnoreCase(s2last) &&

isReverse(s1.substring(1), s2.substring(0, s2.length() - 1));

}

}

public static boolean isReverse(String s1, String s2)

{

if (s1.length() != s2.length()) {

return false; // not same length

} else if (s1.length() == 0 && s2.length() == 0) {

return true;

} else

{

s1 = s1.toLowerCase();

s2 = s2.toLowerCase();

return s1.charAt(0) == s2.charAt(s2.length() - 1) &&

isReverse(s1.substring(1, s1.length()),

s2.substring(0, s2.length() - 1));

}

}

public static boolean isReverse(String s1, String s2)

{

if (s1.length() == s2.length())

{

return isReverse(s1.toLowerCase(), 0, s2.toLowerCase(), s2.length() - 1);

} else {

return false;

// not same length

}

}

private static boolean isReverse(String s1, int i1, String s2, int i2)

{

if (i1 >= s1.length() && i2 < 0) {

return true;

} else {

return s1.charAt(i1) == s2.charAt(i2) &&

isReverse(s1, i1 + 1, s2, i2 - 1);

}

}

public static boolean isReverse(String s1, String s2)

{

return reverse(s1.toLowerCase()).equals(s2.toLowerCase());

}

private static String reverse(String s)

{

if (s.length() == 0) {

return s;

} else {

return reverse(s.substring(1)) + s.charAt(0);

}

}

Explanation:

building and maintaining the topology is difficult and time consuming. the chance of redundant connections is high, which adds to the high costs and potential for reduced efficiency.

explanation: