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copied!<p>Probably just me, but I don't like any of the solutions provided.</p> <pre><code> public static LinkedList Of(String[] input) { if (input == null || input.length < 1) return null; return LinkedList.Of(input, input.length - 1); } public static LinkedList Of(String[] input, int i) { if (i > 0) return new LinkedList(input[i], LinkedList.Of(input, i - 1)); return new LinkedList(input[i], null); } </code></pre> <p>Down vote is scary, so here is it in the right order:</p> <pre><code> /** * Creates linked list from array input. * * @param input * data array * @return linked list with data */ public static LinkedList Of(String[] input) { // Checks if array has elements. if (input == null || input.length < 1) return null; // Starts creating the array using overload 2. return LinkedList.Of(input, 0); } /** * Creates linked list from array input (overload 2). * * @param input * data array * @param i * counter to remember at what element is current * @return linked list with data */ public static LinkedList Of(String[] input, int i) { //Tests if counter is within array elements. if (input.length - 1 > i) // Returns new element with (current element data, reference // to next element). Note that next element will be returned // by this same method (this is why it is recursive). return new LinkedList(input[i], LinkedList.Of(input, i + 1)); //Last element. From here backtracking will begin. return new LinkedList(input[i], null); } </code></pre> <p>Here is something for up vote:</p> <pre><code> public String toString() { StringBuilder sb = new StringBuilder(this.word); LinkedList tmp = this; while (tmp.next != null) { sb.append(" > "); tmp = tmp.next; if (tmp.word != null) sb.append(tmp.word); } return sb.toString(); } </code></pre> <p>And to test:</p> <pre><code> String str = "Neque porro quisquam est qui dolorem ipsum quia " + "dolor sit amet, consectetur, adipisci velit..."; LinkedList ll = LinkedList.Of(str.split("\\s+")); System.out.println(ll); </code></pre>

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