Add support for the multiple list in $.chain().concat().

Author: javadev

src/main/java/com/github/underscore/$.java

@@ -1752,8 +1752,8 @@ public <F> Chain<F> uniq(final Function1<T, F> func) {
             return new Chain<F>($.newArrayList((Iterable<F>) $.uniq(list, func)));
         }
 
-        public Chain<T> concat(final List<T> second) {
-            return new Chain<T>($.concat(list, second));
+        public Chain<T> concat(final List<T> ... lists) {
+            return new Chain<T>($.concat(list, lists));
         }
 
         public <T> Chain<T> slice(final int start) {

src/test/java/com/github/underscore/UnderscoreTest.java

@@ -90,6 +90,7 @@ public void concat() {
         assertEquals(asList(1, 2, 3, 4), $.concat(asList(1, 2), asList(3, 4)));
         assertEquals(asList(1, 2, 3, 4), new $(asList(1, 2)).concatWith(asList(3, 4)));
         assertEquals("[1, 2, 3, 4]", $.chain(asList(1, 2)).concat(asList(3, 4)).value().toString());
+        assertEquals("[1, 2, 3, 4, 5, 6]", $.chain(asList(1, 2)).concat(asList(3, 4), asList(5, 6)).value().toString());
         assertEquals(asList(1, 2, 3, 4), asList($.concat(new Integer[] {1, 2}, new Integer[] {3}, new Integer[] {4})));
         assertEquals(asList(1, 2, 3, 4), $.concat(asList(1, 2), asList(3), asList(4)));
         assertEquals(asList(1, 2, 3, 4), new $(asList(1, 2)).concatWith(asList(3), asList(4)));