CIS 265 Exam 2 First Name Last Name

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1 CIS 265 Exam 2 First Name Last Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Which of the data types below does not allow duplicates? 1) A) List B) Set C) LinkedList D) Vector E) Stack 2) Which of the data types below could be used to store elements in their natural order based on the compareto method. A) LinkedHashSet B) Set C) Collection D) TreeSet E) HashSet 2) 3) To get an iterator from a set, you may use the method. 3) A) iterator B) finditerator C) getiterator D) iterators 4) Suppose set s1 is [1, 2, 5] and set s2 is [2, 3, 6]. After s1.addall(s2), s1 is. 4) A) [1, 5] B) [1, 2, 3, 5, 6] C) [2] D) [1, 2, 2, 3, 5, 6] 5) Suppose set s1 is [1, 2, 5] and set s2 is [2, 3, 6]. After s1.retainall(s2), s1 is. A) [1, 5] B) [1, 2, 2, 3, 5, 6] C) [1, 2, 3, 5, 6] D) [2] 5) 6) Suppose set s1 is [1, 2, 5] and set s2 is [2, 3, 6]. After s1.addall(s2), s2 is. 6) A) [2, 3, 6] B) [1, 5] C) [1, 2, 3, 5, 6] D) [1, 2, 2, 3, 5, 6] E) [2] 7) Suppose set s1 is [1, 2, 5] and set s2 is [2, 3, 6]. After s1.removeall(s2), s1 is. A) [1, 5] B) [1, 2, 3, 5, 6] C) [2] D) [1, 2, 2, 3, 5, 6] 8) The printout of the following code is. 7) 8) LinkedHashSet<String> set1 = new LinkedHashSet<String>(); set1.add("new York"); LinkedHashSet<String> set2 = (LinkedHashSet<String>)(set1.clone()); set1.add("atlanta"); set2.add("dallas"); System.out.println(set2); A) [New York, Dallas] B) [New York, Atlanta] C) [New York, Atlanta, Dallas] D) [New York]

2 9) Which of the following is correct to perform the set intersection of two sets s1 and s2? A) s1.intersection(s2) B) s1.retainall(s2) C) s1.join(s2) D) s1.intersect(s2) 10) Analyze the following code: 9) 10) public class Test { public static void main(string[ ] args) { Map map = new HashMap(); map.put("123", "John Smith"); map.put("111", "George Smith"); map.put("123", "Steve Yao"); map.put("222", "Steve Yao"); A) A runtime error occurs because two entries with the same key "123" are added to the map. B) After all the four entries are added to the map, "123" is a key that corresponds to the value "Steve Yao". C) After all the four entries are added to the map, "John Smith" is a key that corresponds to the value "123". D) After all the four entries are added to the map, "123" is a key that corresponds to the value "John Smith". E) After all the four entries are added to the map, "Steve Yao" is a key that corresponds to the value "222".

3 (40 points) Java Programming Write a Java method that returns a new list holding its string data in size and alphabetical order. The method accepts for input a list holding (non-duplicated) strings in an arbitrary order. Your solution produces a new list that places the input strings elements in (FIRST) increasing size order, and (SECOND) alphabetical order in each size group. As an example, if the input list holds [ xx, zzz, bb, aaaa, ddd ] the output list should be [ bb, xx, ddd, zzz, aaaa ]. HINT: This is similar to the problem of tallying words from a file. Consider the following suggestion. Sort alphabetically the list, create a map in which the size of each string is its key and the strings are the value. Traverse the map combining the values. Do the appropriate conversion to ArrayList. Return results. The header of the method is given below public static ArrayList<String> orderbysize(arraylist<string> list)

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5 1) B 2) D 3) A 4) B 5) D 6) A 7) A 8) A 9) B 10) B public class Driver { public static void main(string[] args) { ArrayList<String> originallist = new ArrayList<String> (Arrays.asList("xx", "zzz", "bb", "aaaa", "ddd")); System.out.println ( originallist ); //main ArrayList<String> sortedlist = orderbysize( originallist ); System.out.println ( sortedlist ); for(int i=0; i<sortedlist.size() ; i++) { System.out.printf ( "\n %d: %s", i, sortedlist.get(i) ); private static ArrayList<String> orderbysize(arraylist<string> originallist) { ArrayList<String> solution = originallist; Collections.sort( solution ); System.out.println (solution); TreeMap<Integer, String> map = new TreeMap<Integer, String>(); Integer key; String value; for (int i=0; i<solution.size(); i++ ){ key = solution.get(i).length(); value = solution.get(i); if ( map.get(key) == null ){ //first time - no other string of this size here map.put(key, value); else { value = map.get(key) + " " + value; map.put(key, value); System.out.println ( map ); key = map.firstkey(); String allstrings = ""; while( key!= null){ allstrings += map.get(key) + " "; key = map.higherkey(key); System.out.println ( "allstrings>> " + allstrings ); return new ArrayList<String>(Arrays.asList(allStrings.trim().split(" ")));

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