J2EE+招聘試題.doc

網龍筆試題：1、實現鏈表的創(chuàng)建，刪除和逆序等。#include#includeusing namespace std;typedef struct stuint data;struct stu *next;node, list;node* creatlist(int n)node *h, *p, *s;h = new node;h-next = NULL;p = h;cout “please input ” n ” numbers.”;for(int i = 0; i s-data;s-next = NULL;p-next = s;p = s;return h;void deletelist (node *s, int a)node *p;while(s-data != a)p = s;s = s-next;if(s = NULL)cout next = s-next;delete s;void display(node *h)h = h-next;while(h != NULL)cout data next;cout next;q = p-next;while(q-next != NULL)s = q-next;q-next = p;p = q;q = s;q-next = p;h-next-next = NULL;h-next = q;return h;int main()int n;node *h;cout n;h= creatlist(n);display(h);cout n;deletelist(h, n);display(h);h = reverse(h);display(h);return 0;2、字符串逆序輸入文件是in.txt，內容例如如下：hello, everyone.my name is shen, he.ni, ne?則倒序之后為：everyone. hello,he. shen, is name myne? ni,源程序，自己實現的，算法是別人的。呵呵#include#include#include#include#includeusing namespace std;int main()ifstream inFile;ofstream outFile;char *fName;string:size_type b,e;char ch;string line;vector svec;inFile.open(”in.txt”);while(getline(inFile, line) /先獲取每一行字符，并存入容器中svec.push_back(line);inFile.close();outFile.open(”out.txt”);vector:iterator iter = svec.begin();while(iter != svec.end()b = 0;e = (*iter).size() 1;while( b e) /將該行字符整體翻轉ch = (*iter)b;(*iter)b = (*iter)e;(*iter)e = ch;b+;e;b = 0;e = 0;string:size_type i = 0;while( (*iter)i )if(*iter)i != ) /找到一行的一個單詞b = i;while(*iter)i != & (*iter)i)i+;i;e = i;while (b e) /將這個單詞翻轉ch = (*iter)b;(*iter)b = (*iter)e;(*iter)e = ch;b+;e;i+;outFile *iter endl;+iter;return 0;新大陸面試題：(一)技術面試struts如何表現MVC模式的？Tomcat的部署?UML有哪些圖?Struts框架的好處(優(yōu)點)?答：1、是開源軟件，使開發(fā)者能更深入的了解其內部實現機制。 2、Taglib是Struts的標記庫，靈活動用，能大大提高開發(fā)效率；通過一個配置文件，即可把握整個系統(tǒng)各部分之間的聯系，這對于后期的維護有著莫大的好處。尤其是當另一批開發(fā)者接手這個項目時，這種優(yōu)勢體現得更加明顯。 4. 提供Exception處理機制 . 5. 數據庫鏈接池管理 6. 支持I18N Struts+Hibernate框架的優(yōu)勢?答：這兩種架構相結合很好地解決了系統(tǒng)的開發(fā)效率低、不易于維護、低耦合及可移植性差等問題.介紹一下AJAX?答：AJAX(Asynchronous JavaScript and XML)被贏得廣泛的認可，其原因是由于它縮短了Web應用程序和桌面應用程序之間的差距，并在其中充分結合可實現的技術和豐富的用戶體驗。AJAX是多種技術的綜合，它打破了頁面刷新的范式，使您的用戶快速方便的與 Web 應用程序交互。CVS配置與使用？POJO是什么?答：簡單的Java對象（Plain Old Java Objects），POJO對象有時也被稱為Data對象，大量應用于表現現實中的對象。概要設計與詳細設計等文檔的主要包含內容？(二)筆試內容Java基礎:Char變量.ClassLoader與Class.JDO.GC.等SQL方面:SQL全稱.查詢語句. Group by .having.update setJSP方面:靜態(tài)include與動態(tài)include第一部分: Java 基礎知識1. What will happen when you attempt to compile and run the following code? public class Test static int x = 5; static int x, y; public static void main(String args) x-; myMethod(); System.out.println(x + y + +x); public static void myMethod() y = x+ + +x; A. compiletime error B. prints: 1 C. prints: 2 D. prints: 3 E. prints: 7 F. prints: 8 2. Which of the following collection classes from java.util package are Thread safe? (Choose two)A. Vector B. ArrayListC. HashMap D. Hashtable 3 Which of the following types is primitive java type？(Choose two)A StringB intC charD Short4Which two demonstrate a has a relationship? (Choose two) A. public interface Person public class Employee extends Person B. public interface Shape public interface Rectandle extends Shape C. public interface Colorable public class Shape implements Colorable D. public class Species public class Animalprivate Species species; E. interface Component class Container implements Component private Component children; 5. What is the result when you compile and run the following code? public class ThrowsDemo static void throwMethod() System.out.println(Inside throwMethod.); throw new IllegalAccessException(demo); public static void main(String args) try throwMethod(); catch (IllegalAccessException e) System.out.println(Caught + e); A. compile error B. runtime error C. compile successfully, nothing is printed. D. inside throwMethod followed by caught: java.lang.IllegalAccessException: demo 6. Which two statements are true for the class java.util.TreeSet? (Choose two) A. The elements in the collection are ordered. B. The collection is guaranteed to be immutable. C. The elements in the collection are guaranteed to be unique. D. The elements in the collection are accessed using a unique key. E. The elements in the collection are guaranteed to be synchronized 點評：TreeSet類實現了Set接口。Set的特點是其中的元素惟一，選項C正確。由于采用了樹形存儲方式，將元素有序地組織起來，所以選項A也正確。7. What will be printed when you execute the following code? class X Y b = new Y(); X() System.out.print(X); class Y Y() System.out.print(Y); public class Z extends X Y y = new Y(); Z() System.out.print(Z); public static void main(String args) new Z(); A. Z B. YZ C. XYZ D. YXYZ 8. which two declaretions prevent the overriding of a method? A. final void methoda() B. void final methoda() C. static void methoda() D. static final void methoda() E. final abstract void methoda() 9. You want a class to have access to members of another class in the same package which is the most restrictive access modifier that will accomplish this objective? A. public B. private C. protected D. transient E. No access modifier is required 10. which two interfaces provide the capability to store objects using a key-value pair? A. java.util.Map B. java.util.Set C. java.util.List D. java.util.SortedSet E. java.util.SortedMap F. java.util.Collection 11.1) class Super 2) public float getNum()return 3.0f; 3) 4) 5) public class Sub extends Super 6) 7) which method, placed at line 6, will cause a compiler error? A. public float getNum()return 4.0f; B. public void getNum() C. public void getNum(double d) D. public double getNum(float d)return 4.0d; 注意這道題主要考的是方法的overload和override。對于overload，只有參數列表不同，才做為標準，而返回值和訪問控制關鍵字不能做為標準，所以B錯在方法名相同，但只有返回值不同，這是錯的。C和D是正確的overload。對于override，則訪問控制關鍵字只能更加公有化，異常只能是超類方法拋出的異常的子類，也可以不拋出。返回類型，參數列表必須精確匹配。所以A是正確的override。12. which gets the name of the parent directory of file file.txt? A. String name=File.getParentName(file.txt); B. String name=(new File(file.txt).getParent(); C.Stringname=(new File(file.txt).getParentName(); D.Stringname=(new File(file.txt).getParentFile(); E.Diretorydir=(new File(file.txt).getParentDir(); String name=dir.getName(); 13What happens when you try to compile and run the following application? Choose all correct options. 1. public class Z 2. public static void main(String args) 3. new Z(); 4. 5. 6. Z() 7. Z alias1 = this; 8. Z alias2 = this; 9. synchronized(alias1) 10. try 11. alias2.wait(); 12. System.out.println(“DONE WAITING”); 13. 14. catch (InterruptedException e) 15. System.out.println(“INTERR UPTED”); 16. 17. catch (Exception e) 18. System.out.println(“OTHER EXCEPTION”); 19. 20. finally 21. System.out.println (“FINALLY”); 22. 23. 24. System.out.println(“ALL DONE”); 25. 26. A. The application compiles but doesnt print anything. B. The application compiles and print “DONE WAITING” C. The application compiles and print “FINALLY” D. The application compiles and print “ALL DONE” E. The application compiles and print “INTERRUPTED” 點評：在Java中，每一個對象都有鎖。任何時候，該鎖都至多由一個線程控制。由于alias1與alias2指向同一對象Z，在執(zhí)行第11行前，線程擁有對象Z的鎖。在執(zhí)行完第11行以后，該線程釋放了對象Z的鎖，進入等待池。但此后沒有線程調用對象Z的notify()和notifyAll()方法，所以該進程一直處于等待狀態(tài)，沒有輸出。14. What results from attempting to compile and run the following code? public class Ternary public static void main(String args) int a = 5; System.out.println(Value is - + (a 5) ? 9.9 : 9); A. print:Value is -9 B. print:Value is -5 C. Compilation error D. None of these第二部