先总体说下题型,共有20道选择题,4道简答题,3道编程题和1道扩展题,题目都比较简单,限时一小时完成。
一、选择题
选择题非常简单,都是基础题,什么死锁发生的条件、HashMap和HashSet查找插入删除的时间复杂度、Thread类和Runnable接口、排序复杂度比较、建堆调整堆等等,具体的也记不得了。
二、简答题
1. 简述Servlet的生命周期
2. 写出至少8个Java常用的包名称
3. Overload和Override的区别,Overloaded方法能不能修改返回值类型?
4. 不用中间变量交换a和b的值
三、编程题
1. 有N个人围一圈依次报数,数到3的人出列,问当只剩一个人时他原来的位子在哪里?
2. 有两个已递增有序的单链表pLinkList和qLinkList,将这两个链表合并成一个递增有序的链表,请自己定义单链表的结构。
3. 具体题目不记得,大概意思就是:从N个数中随机抽取出M个数(M < N),为了使抽取比较均匀,请自己定义抽取函数使得抽取的数既均匀又尽量随机。
四、扩展题
具体题目也记不清了,一大堆,大概意思是:有一个海量日志库,里面的每条日志记录都有相应的关键词和访问次数,但记录是无序的,为了挖掘客户偏好,需要找出前N个最高访问次数的日志记录,请设计算法尽量使时间复杂度和空间复杂度最低。
下面是我自己写的答案,不一定正确,欢迎大家批评指定和提出自己更好的想法和意见:
1. 简述Servlet的生命周期
答:Web容器加载servlet,生命周期开始,通过调用servlet的的init()方法进行servlet的初始化,通过调用service()方法实现,根据请求的不同调用不同的doGet()和doPost()方法,结束服务,web容器调用servlet的destroy()方法。
一个servlet的生命周期由部署servlet的容器控制,当一个请求映射到一个servlet时,容器执行下步骤:
1.加载servlet类
2.创建一个servlet类的实例
3.调用init初始化servlet实例,
4.调用service方法,传递一个请求和响应对象
5.容器要移除一个servlet,调用servlet的destroy方法结束该servlet
2. 写出至少8个Java常用的包名称
答:答出以下的任意8个就行了
1. class="Apple-converted-space"> java.lang Java 编程语言的基本类库
2. java.applet 创建 applet 需要的所有类
3. java.awt 创建用户界面以及绘制和管理图形、图像的类
4. java.io 通过数据流、对象序列以及文件系统实现的系统输入、输出
5. java.net 用于实现网络通讯应用的所有类
6. java.util 集合类、时间处理模式、日期时间工具等各类常用工具包
7. java.sql 访问和处理来自于 Java 标准数据源数据的类
8. java.test 以一种独立于自然语言的方式处理文本、日期、数字和消息的类和接口
9. java.security 设计网络安全方案需要的一些类
10. java.beans 开发 Java Beans 需要的所有类
11. java.math 简明的整数算术以及十进制算术的基本函数
12. java.rmi 与远程方法调用相关的所有类
3. Overload和Override的区别,Overloaded方法是否可以改变返回值类型?
答:Overload是重载的意思,Override是覆盖的意思,也就是重写。
(1)重载Overload表示同一个类中可以有多个名称相同的方法,但这些方法的参数列表各不相同(即参数个数或类型不同),重载发生在同一个类中。
(2)重写Override表示子类中的方法可以与父类中的某个方法的名称和参数完全相同,通过子类创建的实例对象调用这个方法时,将调用子类中的定义方法,这相当于把父类中定义的那个完全相同的方法给覆盖了,这也是面向对象编程的多态性的一种表现。子类覆盖父类的方法时,只能比父类抛出更少的异常,或者是抛出父类抛出的异常的子异常,因为子类可以解决父类的一些问题,不能比父类有更多的问题。子类方法的访问权限只能比父类的更大,不能更小。如果父类的方法是private类型,那么,子类则不存在覆盖的限制,相当于子类中增加了一个全新的方法。重写发生在不同的类(父类和子类)中。
(3)至于Overloaded的方法是否可以改变返回值的类型这个问题,要看你倒底想问什么呢?这个题目很模糊。如果几个Overloaded的方法的参数列表不一样,它们的返回者类型当然也可以不一样。但我估计你想问的问题是:如果两个方法的参数列表完全一样,是否可以让它们的返回值不同来实现重载Overload。这是不行的,我们可以用反证法来说明这个问题,因为我们有时候调用一个方法时也可以不定义返回结果变量,即不要关心其返回结果,例如,我们调用map.remove(key)方法时,虽然remove方法有返回值,但是我们通常都不会定义接收返回结果的变量,这时候假设该类中有两个名称和参数列表完全相同的方法,仅仅是返回类型不同,java就无法确定编程者倒底是想调用哪个方法了,因为它无法通过返回结果类型来判断。
4. 不用中间变量交换a和b的值
答:很多种方法,我这里给出最简单的:
a = a + b;
b = a - b;
a = a - b;
1. 有N个人围一圈依次报数,数到3的倍数的人出列,问当只剩一个人时他原来的位子在哪里?
解答:经典的转圈踢人问题,好吧专业一点,约瑟夫环问题,相信大家都会,下面给我的code:
int main() { int N, i, j; printf("Please enter the number of people(N): "); scanf("%d", &N); int *pArray = (int *)malloc(sizeof(int) * N); int count = 0; // 这里编号为0 ~ N - 1 for(i = 0; i < N; i++) { pArray[i] = i; } for(i = 0, j = 0; i < N; i = (i + 1) % N) { if(pArray[i] != -1) { j++; if(j % 3 == 0) { pArray[i] = -1; count++; if(count == N) { printf("The last people is %d\n", i); break; } } } } return 0; }
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alt="" />
好吧,我承认我的算法很臃肿,完全是模拟了整个游戏过程,时间复杂度为O(mn),这里m=3,网上有个大牛给出了归纳数学的方法,具体方法如下:
为了讨论方便,先把问题稍微改变一下,并不影响原意:
问题描述:n个人(编号0~(n-1)),从0开始报数,报到(m-1)(这里m=3)的退出,剩下的人继续从0开始报数,求最后剩下一个人的编号。
我们知道第一个人(编号一定是m%n-1) 出列之后,剩下的n-1个人组成了一个新的约瑟夫环(以编号为k=m%n的人开始):
k k+1 k+2 ... n-2, n-1, 0, 1, 2, ... k-2并且从k开始报0。
现在我们把他们的编号做一下转换:
k --> 0
k+1 --> 1
k+2 --> 2
...
...
k-2 --> n-2
k-1 --> n-1
变换后就完完全全成为了(n-1)个人报数的子问题,假如我们知道这个子问题的解:例如x是最终的胜利者,那么根据上面这个表把这个x变回去不刚好就是n个人情况的解吗?!!变回去的公式很简单,相信大家都可以推出来:x'=(x+k)%n
如何知道(n-1)个人报数的问题的解?对,只要知道(n-2)个人的解就行了。(n-2)个人的解呢?当然是先求(n-3)的情况 ---- 这显然就是一个倒推问题!好了,思路出来了,下面写递推公式:
令f[i]表示i个人玩游戏报m退出最后胜利者的编号,最后的结果自然是f[n]
递推公式
f[1]=0;
f[i]=(f[i-1]+m)%i; (i>1)
有了这个公式,我们要做的就是从1-n顺序算出f[i]的数值,最后结果是f[n]。因为实际生活中编号总是从1开始,我们输出f[n]+1
由于是逐级递推,不需要保存每个f[i],程序也是异常简单:
#include <stdio.h> #include <stdlib.h> int main() { int N, i, s = 0; printf("Please enter the number of people(N): "); scanf("%d", &N); for (i = 2; i <= N; i++) { s = (s + 3) % i; } printf ("The last people is %d\n", s); return 0; }
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" alt="" />
这个算法的时间复杂度为O(n),相对于模拟算法已经有了很大的提高。算n,m等于一百万,一千万的情况不是问题了。可见,适当地运用数学策略,不仅可以让编程变得简单,而且往往会成倍地提高算法执行效率。数学确实很重要啊!!!
2. 有两个已递增有序的单链表pLinkList和qLinkList,将这两个链表合并成一个递增有序的链表,请自己定义单链表的结构。
解答:同样很经典,不用多说,直接上我自己的code(不是最好的):
#include <iostream> using namespace std; struct LinkList { int data; LinkList *next; }; LinkList* createList() { LinkList *head = NULL, *p, *q; int data; cin >> data; while(data) { p = new LinkList; p->data = data; p->next = NULL; if(head == NULL) { head = p; q = head; } else { q->next = p; q = p; } cin >> data; } return head; } // 合并后的链表放在pLinkList中 void merge(LinkList *&pLinkList, LinkList *qLinkList) { LinkList *pre, *p, *q; pre = NULL; p = pLinkList; q = qLinkList; while(p != NULL && q != NULL) { if(p->data < q->data) { pre = p; p = p->next; } else { // 如果p第一个结点大于q,则改变合并后头结点为q if(pre == NULL) { pLinkList = q; } else { pre->next = q; } pre = q; q=q->next; pre->next = p; } } // 最后不要忘了qLinkList剩余的大结点 if(q != NULL) { pre->next = q; } } void print(LinkList *l) { LinkList *p = l; while(p != NULL) { if(p->next == NULL) { cout << p->data; break; } cout << p->data << " -> "; p = p->next; } cout << endl; } int main() { cout << "Please enter pLinkList: "; LinkList *pLinkList = createList(); print(pLinkList); cout << "\nPlease enter pLinkList: "; LinkList *qLinkList = createList(); print(qLinkList); merge(pLinkList, qLinkList); cout << "\nThe merge LinkList is: \n"; print(pLinkList); return 0; }
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3. 具体题目不记得,大概意思就是:从N个数中随机抽取出M个数(M < N),为了使抽取比较均匀,请自己定义抽取函数使得抽取的数既均匀又尽量随机。
解答:当时时间太急了,没来得及多想,做法很傻:从1 ~ M*N中随机抽取一个数字,然后mod (N + 1),求得的值为N个数中的下标,再根据此下标去N个数中取,重复M次即可。假如这N个数存在数组nArray[]中,抽取的M个数存在数组mArray[]中,伪代码描述如下:
for(int i = 0; i < M; i++) { int index = Random(M * N) % N; mArray[i] = nArray[index]; }
由于觉得这个算法实在是不好,就懒得测试了,大家有好想法的赶紧提出来吧。
具体题目也记不清了,一大堆描述,大概意思是:有一个海量日志库,里面的每条日志记录都有相应的关键词和访问次数,但记录是无序的,为了挖掘客户偏好,需要找出前N个最高访问次数的日志记录,请设计算法尽量使时间复杂度和空间复杂度最低。
解答:典型的Top K问题,我用的算法也是大家都知道的,大致描述下思路:假如关键词和访问次数成一个记录结构体,维护一个有N个该结构体的小根堆,初始化为N个日志记录的关键词和访问次数(建堆算法),每次有新的记录时,将该记录的访问次数与小根堆的堆顶元素进行比较,如果大于堆顶元素则与堆顶元素交换记录,然后调整堆结构使其重新为一个小根堆,否则置之不理。当所有记录遍历完后,所有的堆元素就是所要求的前N个最高访问次数的日志记录。时间复杂度为O(MlgN),不知道自己分析的对不对,完全是自以为是的想法,如果大家有更好的算法欢迎提出!