Q.1
public static Collection get() {
Collection sorted = new LinkedList();
sorted.add("B"); sorted.add("C"); sorted.add("A");
return sorted;
}
public static void main(String[] args) {
for(Object obj : get()) {
System.out.print(obj+", ");
}
}
What is the result?
A. A, B, C,
B. B, C, A,
C. Compilation fails
D. The code runs with no output
E. An exception is thrown at runtime
Ans: B
因為沒有new Treeset 所以不會排序還是照資料進入順序印出
也可以透過TreeSet的Constructor排序
但資料本身必須實作comparable介面，資料才有辦法運作
Q.2
import java.util.*;
public class Test {
enum Example { ONE, TWO, THREE }
public static void main(String[] args) {
Collection coll=new ArrayList();
coll.add(Example.THREE);
coll.add(Example.THREE);
coll.add(Example.THREE);
coll.add(Example.TWO);
coll.add(Example.TWO);
coll.add(Example.ONE);
Set set=new HashSet(coll);
}
}
Which statement is true about the set variable in main method ?
variable 變數
contains 包含
elements 元素, 資料
order 順序
preserved 保留
A. The set variable contains all six elements from the coll collection, and the order is guaranteed to be preserved.
B. The set variable contains only three elements from the coll collection, and the order is guaranteed to be preserved.
C. The set variable contains all six elements from the coll collection, but the order is NOT guaranteed to be preserved.
D. The set variable contains only three elements from the coll collection, but the order is NOT guaranteed to be preserved.
My ans: C
答案: D
因為剛好是enum HashCode, Equals剛好有符合特性 記憶體位置相同 必須被覆寫 為何enum可以自動過濾?
Q.3
import java.util.*;
public class Example {
public static void main(String[] args) {
//insert code here
set.add(new Integer(2));
set.add(new Integer(1));
System.out.println(set);
}
}
Which code, inserted at "//insert code here", guarantees that this program will output [1, 2]?
A. Set set = new TreeSet();
B. Set set = new HashSet();
C. Set set = new SortedSet();
D. List set = new SortedList();
E. Set set = new LinkedHashSet();
My ans: A 用TreeSet自動排序
答案:A. D是亂寫的 沒用
Q.4
34. HashMap props=new HashMap();
35. props.put("key45", "some value");
36. props.put("key12", "some other value");
37. props.put("key39", "yet another value");
38. Set s = props.keySet();
39. // insert code here
What, inserted at line 39, will sort the keys in the props HashMap?
A. Arrays.sort(s);
B. s = new TreeSet(s);
C. Collections.sort(s);
D. s =new SortedSet(s);
My ans: C
答案: B, 用TreeSet()就對了。C錯因為參數數量不對，而且____
Q.5
Given :
public class Key {
private long id1;
private long id2;
// class key methods
}
A programmer is developing a class Key, that will be used as a key in a standard java.util.HashMap. Which two methods should be overridden to assure that key works correctly as a key? (choose two)
standard 標準
assure 保證
works 工作
correctly 正確地
A. public int hashCode ( )
B. public boolean equals (Key k)
C. public int compareTo (Object o)
D. public boolean equals (Object o)
My ans: AB
答案: AD. C是用排序與本題無關 B錯因為要付血必須與父類別名稱完全一樣(equals的父類別是Object)
Q.6
public class Person {
private String name, comment;
private int age;
public Person(String n, int a, String c) {
name=n;
age=a;
comment=c;
}
public boolean equals(Object o) {
if(! (o instanceof Person)) return false;
Person p = (Person)o;
return age==p.age && name.equals(p.name);
}
}
What is the appropriated definition of the hashCode method in class Person?
definition 定義
A. return super.hashCode();
B. return name.hashCode() + age * 7;
C. return name.hashCode() + comment.hashCode() / 2;
D. return super.hashCode() + comment.hashCode() / 2 - age * 3;
答案: B 符合雜湊碼原則
Q.7
1. public class Person {
2. private String name;
3. public Person(String name) { this.name=name; }
4. public boolean equals(Person p) {
5. return p.name.equals(this.name);
6. }
7. }
Which statement is true?
properly 恰當地, 正確地
attribute 屬性
prevent 防止
duplicates 重複
A. The equals method does NOT properly override the Object.equals method.
B. Compilation fails because the private attribute p.name cannot be accessed in line 5.
C. To work correctly with hash-based data structures, this class must also implement the hashCode method.
D. When adding Person objects to java.util.Set collection, the equals method in line 4 will prevent duplicates.
答案: A 因為equals沒有辦法正常被覆寫，第4行寫錯了。C是正確，但仍需要修正第四4行。
Q.8
public static Iterator reverse(List list) {
Collections.reverse(list);
return list.iterator();
}
public static void main(String[] args) {
List list=new ArrayList();
list.add("1"); list.add("2"); list.add("3");
for(Object obj : reverse(list))
System.out.print(obj+", ");
}
What is the result?
A. 3, 2, 1,
B. 1, 2, 3,
C. Compilation fails
D. The code runs with no output
E. An exception is thrown at runtime
答案: C, list不收iterator
Q.9
import java.util.*;
public class LetterASort {
public static void main(String[] args) {
ArrayList<String> strings=new ArrayList<String>();
strings.add("aAaA");
strings.add("AaA");
strings.add("aAa");
strings.add("AAaa");
Collections.sort(strings);
for(String s : strings) { System.out.print(s+" "); }
}
}
What is the result?
A. Compilation fails
B. aAaA aAa AAaa AaA
C. AAaa AaA aAa aAaA
D. AaA AAaa aAaA aAa
E. aAa AaA aAaA AAaa
F. An exception is thrown at runtime
答案: C, 照unicode順序排序, 所以小a在前,大A在後，筆順序比長短。E一定不會出現。
Q.10
import java.util.*;
public class WrappedString {
private String s;
public WrappedString(String s) { this.s=s; }
public static void main(String[] args) {
HashSet<Object> hs=new HashSet<Object>();
WrappedString ws1=new WrappedString("aardvark");
WrappedString ws2=new WrappedString("aardvark");
String s1=new String("aardvark");
String s2=new String("aardvark");
hs.add(ws1); hs.add(ws2); hs.add(s1); hs.add(s2);
System.out.println(hs.size());
}
}
What is the result?
A. 0
B. 1
C. 2
D. 3
E. 4
F. Compilation fails.
G. An exception is thrown at runtime
答案:D, 有Hash=>用雜湊碼，必須覆寫equals, HashCode, 有Set=>不會有重覆值
ws1,2 對系統來看是覆寫最高父類別的equals, 所以存的是記憶體位置，而ws1.2記憶體位置並不同,因此兩筆都會被記錄。String的話有覆寫equals 和hascode,因此比較的是實際內容不是記憶體位置,所以HashSet的公用可以發揮。總共會有ws1,ws2,s2共三筆。
Q.11
Given the exhibit:
import java.util.*;
public class PQ {
public static void main(String[] args) {
PriorityQueue<String> pq=new PriorityQueue<String>();
pq.add("carrot");
pq.add("apple");
pq.add("banana");
System.out.println(pq.poll()+":"+pq.peek());
}
}
What is the result?
A. apple:apple
B. carrot:apple
C. apple:banana
D. banana:apple
E. carrot:carrot
F. carrot:banana
答案: C. PriorityQueue還是會用自然排序，所以變成apple, banana, carrot. 然後poll是取第一筆資料然後移除, peek是取第一筆資料然後不宜出。所以System.out.println(pq.poll()+":"+pq.peek());會印出apple:banana
Q.12
// insert code here
private N min, max;
public N getMin() { return min; }
public N getMax() { return max; }
public void add(N added) {
if(min==null || added.doubleValue() < min.doubleValue())
min=added;
if(max==null || added.doubleValue() < max.doubleValue())
max=added;
}
}
Which two, inserted at line 1 will allow the code to compile? (Choose Two)
A. public class MinMax< ? > {
B. public class MinMax < ? extends Number> {
C. public class MinMax <N extends Object> {
D. public class MinMax <N extends Number > {
E. public class MinMax < ? extends Object > {
F. public class MinMax < N extends Integer > {
答案: DF, N是佔位符(泛型)。
Q.13
Given:
import java.util.*;
public class Old {
public static Object get0(List list) {
return list.get(0);
}
}
Which three will compile successfully? (Choose three.)
A. Object o = Old.get0(new LinkedList());
B. Object o = Old.get0(new LinkedList<?>());
C. String s = Old.get0(new LinkedList<String>());
D. Object o = Old.get0(new LinkedList<Object>());
E. String s = (String)Old.get0(new LinkedList<String>());
答案: ADE, 需了解何謂<泛型>。
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