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copied!<p>Without more code I can't say for sure, but it is most likely that you're never making <code>series2 = new LinkedList<Number></code> or something similar.</p> <p>Another problem I see is that you're starting at 1. The <code>List</code> interface implementations are typically 0 indexed, so you'd want</p> <pre><code>for(int num = 0; num < series.size(); num += 2) </code></pre> <p>Your best bet is to provide the entire code (what is necessary to compile and reproduce the problem) and the stack trace you're getting on the error as well.</p> <p>EDIT:</p> <p>It also depends on what you want for an output. Let's pretend your input in series is <code>0,1,2,3,4,5,6,7,8,9</code> and that we're dealing with only 10 numbers.</p> <p>Would you want series 2 to be:</p> <ul> <li><code>null,1,null,3,null,5,null,7,null,9</code></li> <li><code>1,null,3,null,5,null,7,null,9,null</code></li> <li><code>1,3,5,7,9</code></li> </ul> <p>If it's one of the first two, then you need to modify your loop to add the extra ones, like so:</p> <pre><code>for (int num = 0; num < 120; num++) { series2.add(num % 2 == 1 ? series.get(num) : null); } </code></pre> <p>Or alternatively:</p> <pre><code>for(int num = 1; num < 120; num+=2) { series2.add(null); series2.add(series.get(num)); } </code></pre> <p><strong>In any event</strong> attempting to add like this: <code>series2.add(num, series.get(num))</code> will probably fail. According to the documentation for <code>List.add(int,E)</code></p> <blockquote> <p>Throws</p> <p>IndexOutOfBoundsException - if the index is out of range (index < 0 || index > size())</p> </blockquote> <p>So you can only use that to add to the middle of the <code>List</code>. If you want to add to the end, you have to use <code>List.add(E)</code>. If you want to add to some specific index in the <code>List</code> that it doesn't contain yet, you need to add <code>null</code>s (or something) until you get there. Obviously this is a problem if your <code>List</code> implementation doesn't support <code>null</code>s.</p>

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