Making space for string (character) in Java

Question

A permutation is a function. In this particulate exercise, the function is certain space separation of an adjacent letter. Here is an example for the space is 2: After the first letter A, skip 2 spaces and arrive at D, skip another 2 space to arrive at G, so original ABC maps to adg.
ABCDEFGHIJKLMNOPQRSTUVWXYZ
maps to (with skip of 2 space):
adgjmpsvybehknqtwzcfilorux
You need to have a user input for the skip space, similar to the lab 1. Warning: Because there are 26 characters in the English alphabetic, if user enters an even number, you need to consider add one every time it passes the Z letter. Please help me to solve this problem.

Answer 1

Try using this code segment:
String abcString = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
StringBuffer tempBuffer = new StringBuffer();
StringBuffer outBuffer = new StringBuffer();
tempBuffer.append(abcString);
i = 0;
while(i < abcString.length())
outBuffer.append(tempBuffer.charAt(i));
i+=2;
tempBuffer.setLength(0);
All you really need to do is change what you're adding to I every time you go through the loop, make it user input. This is simply an example of how to do it.
-Jason Thomas.