Java笔试题

随机从1到100000中间随机取出100个不同的质数，然后按从小到大的顺序排列

public class IsPrime { //工具类
    // 输入一个数,判断是否为质数,费时方法
    public static Integer isPrime(int num) {
        if (num == 0 || num == 1)
            return -1;
        for (int i = 2; i < num - 1; i++) {
            if (num % i == 0) return -1;
        }
        return num;
    }
}
 
 
//        一个大于1的自然数，除了1和它自身外，不能被其他自然数整除的数叫做质数；否则称为合数。
        List<Integer> listrandom=new ArrayList(); //1-1000000数组
        List<Integer> res=new ArrayList(); //所有质数集合
        for(int i=1;i<=100000;i++){
            listrandom.add(i);
        }
        for (Integer integer : listrandom) {
            if((IsPrime.isPrime(integer)!=-1)){
                     res.add(IsPrime.isPrime(integer));
            }
        }
//        int a= (int) Math.floor(Math.random()*res.size());
 
        //100个不同的质数
        List<Integer> listRandom = new ArrayList<Integer>();
        //随机取出n条不重复的数据,这里我设置随机取100条数据
        for (int i = 100; i >=1; i--) {
            Random random = new Random();
            Math.random();
            //在数组大小之间产生一个随机数 j
            int j = random.nextInt(res.size()-1);
            //取得list 中下标为j 的数据存储到 listRandom 中
            listRandom.add(listrandom.get(j));
            //把已取到的数据移除,避免下次再次取到出现重复
            listrandom.remove(j);
        }
//        TreeSet 有序集合 从小到大排序
        List newList = new ArrayList(new TreeSet(listRandom));
        //循环取出 newList 中的数据
        for(Object l:newList) {
            System.out.println(l);
        }

有一个999级的台阶，可以选择每次上1级或者2级，请实现一个方法，计算有多少种上楼的方式。

  public int JumpFloor(int target) {
            if (target == 1){
                return 1;
            }else if(target == 2){
                return 2;
            } else {
                return JumpFloor(target-1)+JumpFloor(target-2);
            }
 
    }

用JSON格式描述下面表格的数据

// json数组
//        static String json = "[
//        {'学号':'1001','姓名':张三,'性别':男,'姓名':26,'年龄':20},
//        {'学号':'1002','姓名':李四,'性别':女,'姓名':26,'年龄':22},
//        {'学号':'1003','姓名':王五,'性别':男,'姓名':26,'年龄':18},
//        ]";

已知两张表完成如下SQL语句：

学生基本信息：
Student
 
ID NUMBER 学号
 
NAME VARCHAR2(10) 姓名
 
SEX VARCHAR2(2) 性别：男女
 
CLASS VARCHAR2(2) 班级：1，2，3，4
 
AGE NUMBER 年龄
 
SUBJECT VARCHAR2(10) 最高分课程
 
SCORE NUMBER 最高成绩
 
 
学生考试成绩：Score
 
ID NUMBER 学号
 
SUBJECT VARCHAR2(10) 课程
 
YEAR NUMBER 考试年份
 
SCORE NUMBER 成绩
 
 
 
 
/*
 Navicat Premium Data Transfer
 Source Server         : 0702
 Source Server Type    : MySQL
 Source Server Version : 50723
 Source Host           : localhost:3306
 Source Schema         : qqq
 Target Server Type    : MySQL
 Target Server Version : 50723
 File Encoding         : 65001
 Date: 13/02/2022 13:04:14
*/
 
SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;
 
-- ----------------------------
-- Table structure for student
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student`  (
  `ID` int(10) NOT NULL COMMENT '学号',
  `NAME` varchar(10) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '姓名',
  `SEX` varchar(2) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '性别：男女',
  `CLASS` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '班级：1，2，3，4',
  `AGE` int(11) NULL DEFAULT NULL COMMENT '年龄',
  `SUBJECT` varchar(10) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '最高分课程',
  `SCORE` int(10) NULL DEFAULT NULL COMMENT '最高成绩',
  PRIMARY KEY (`ID`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
 
-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES (1, '张三', '男', '1', 18, '语文', 90);
INSERT INTO `student` VALUES (2, '李四', '男', '2', 18, '语文', 100);
INSERT INTO `student` VALUES (3, '王五', '男', '2', 18, '语文', 90);
INSERT INTO `student` VALUES (4, '丽丽', '女', '1', 18, '英语', 90);
 
SET FOREIGN_KEY_CHECKS = 1;
 
 
/*
 Navicat Premium Data Transfer
 Source Server         : 0702
 Source Server Type    : MySQL
 Source Server Version : 50723
 Source Host           : localhost:3306
 Source Schema         : qqq
 Target Server Type    : MySQL
 Target Server Version : 50723
 File Encoding         : 65001
 Date: 13/02/2022 13:04:46
*/
 
SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;
 
-- ----------------------------
-- Table structure for score
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score`  (
  `ID` int(10) NULL DEFAULT NULL COMMENT '学号',
  `SUBJECT` varchar(10) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '课程',
  `YEAR` int(255) NULL DEFAULT NULL COMMENT '考试年份',
  `SCORE` int(255) NULL DEFAULT NULL COMMENT '成绩'
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
 
-- ----------------------------
-- Records of score
-- ----------------------------
INSERT INTO `score` VALUES (1, '语文', 2021, 80);
INSERT INTO `score` VALUES (1, '数学', 2021, 90);
INSERT INTO `score` VALUES (1, '英语', 2021, 70);
INSERT INTO `score` VALUES (2, '语文', 2021, 100);
INSERT INTO `score` VALUES (2, '数学', 2021, 70);
INSERT INTO `score` VALUES (4, '英语', 2021, 23);
INSERT INTO `score` VALUES (3, '语文', 2021, 80);
INSERT INTO `score` VALUES (3, '数学', 2021, 12);
INSERT INTO `score` VALUES (3, '英语', 2021, 90);
INSERT INTO `score` VALUES (4, '语文', 2021, 90);
INSERT INTO `score` VALUES (4, '数学', 2021, 45);
INSERT INTO `score` VALUES (2, '语文', 2012, 50);
INSERT INTO `score` VALUES (2, '数学', 2012, 20);
INSERT INTO `score` VALUES (2, '英语', 2012, 30);
 
SET FOREIGN_KEY_CHECKS = 1;

- 列出2012年1班各科成绩的平均值。

SELECT
    AVG( sc.SCORE ) AS '平均成绩',
    sc.YEAR AS '年份',
    st.CLASS AS '班级',
    sc.SUBJECT AS '课程' 
FROM
    score sc
    LEFT JOIN student st ON sc.ID = st.ID 
WHERE
    sc.YEAR = 2021
    AND st.CLASS = 1 
GROUP BY
    sc.SUBJECT 

-- 列出2012年2班数学成绩不及格的学生人员列表。

SELECT
    st.`NAME` AS '姓名',
    sc.YEAR AS '年份',
    st.CLASS AS '班级',
    sc.SUBJECT AS '课程',
    sc.SCORE AS '成绩' 
FROM
    score sc
    LEFT JOIN student st ON sc.ID = st.ID 
WHERE
    sc.YEAR = 2012 
    AND st.CLASS = 2 
    AND sc.SUBJECT = '数学' 
    AND sc.SCORE < 60

-- 查出每位同学成绩所有成绩中分数最高的科目、成绩并更新到学生基本信息中最高分课程和最高成绩字段中。

UPDATE student AS A
INNER JOIN (
SELECT
    MAX( sc.SCORE ) AS maxscore,
    sc.ID AS maxid,
    sc.`SUBJECT` maxsub 
FROM
    score sc
    LEFT JOIN student st ON sc.ID = st.ID 
GROUP BY
    sc.ID 
    ) AS B ON A.ID = B.maxid 
    SET A.ID = B.maxid,
    A.SUBJECT = B.maxsub,
    A.score = B.maxscore

final, finally, finalize 的区别

• final 用于申明属性，方法和类，表示属性不可变，方法不可以被改变，类不可以被继承。
• finally 是异常处理语句结构中，表示总是执行的部分。

• finallize   在垃圾回收器将内存中的对象进行清空之前，允许使用finalize()方法做清理工作

Error 和 Exception 的区别?

1,Exception是程序正常运行中,可以预料的意外情况,可以被捕获,进行相应的处理.

2.Error 是指正常情况下,不大可能出现的情况,绝大部分的Error 都会导致程序处于非正常的,不可恢复的状态, 不需要捕获。

MVC的各个部分都有那些技术来实现

• model： 模型层 应用的业务逻辑（如：数据库的操作），通过JavaBean实现

    （hibernate、mybatis、ibatis）

• view：视图层，用于与用户的交互，主要由jsp页面产生。

• （jsp、FreeMarker、EL、）
• controller：应用层 处理过程控制，一般是一个servlet。

java字符串反转

1、用stringBuffer或者stringBuilder自带的reverse方法

    public static String reverseTestOne(String s) {
        return new StringBuffer(s).reverse().toString();
    }

2.stringBuffer倒序拼接

 public static String reverseTestThree(String s) {
        StringBuffer sb = new StringBuffer();
        for (int i = s.length() - 1; i >= 0; i--) {
            sb.append(s.charAt(i));
        }
        return sb.toString();
    }

3、切割递归反转

  public static String reverseTestSix(String s) {
        if (s.length() <= 1) {
            return s;
        }
        return reverseTestSix(s.substring(1)) + s.substring(0, 1);
    }

4、利用栈的先进后出

   public static String reverseTestFour(String s) {
        StringBuffer sb = new StringBuffer();
        Stack stack = new Stack();
        for (int i = 0; i < s.length(); i++) {
            stack.push(s.charAt(i));
        }
        while (!stack.isEmpty()) {
            //stack会返回栈顶值，并且会把该值删除
            sb.append(stack.pop());
        }
        return sb.toString();
    }

3. SQL间答题

表结构:

1、表名: g_ cardapply

字段(字段

名/类型/长度):

g applynpo varchar 8: /申请单号(关键字)

g_ applydate bigint 8; /申请日期

g_ state varchar 2; //申请状态

2、表名: g_ _cardapplydetail

字段(字段名/类型/长度):

g_ _applyno varchar //申请单号(关键字)

g_ name varchar 30; //申请人姓名

g_ idcard varchar  18; //申请人身份证号

g_ state varchar 2; //申请状态

其中，两个表的关联字段为申请单号。

题目:

-- 1.查询身份证号码为440401430103082的申请日期

SELECT
  a.g_applydate,
  a.g_applyno 
FROM
  g_cardapply a
  LEFT JOIN g_cardapplydetail b ON a.g_applyno = b.g_applyno 
WHERE
  b.g_idcard = '440401430103082'

-- 2.查询同一个身份证号码有两条以上记录的身份证号码以及记录个数

SELECT
    COUNT( 1 ) 
FROM
    ( SELECT * FROM g_cardapplydetail AS s WHERE g_idcard = '440401430103082' ) AS a 
HAVING
    count( 1 ) >= 2

-- 3.将身份证号码为440401430103082的记录在两个表中的申请状态均改为07

​​​​​​​UPDATE g_cardapply a
LEFT JOIN g_cardapplydetail b ON a.g_applyno = b.g_applyno 
SET a.g_state = '1',
b.g_state = '1' 
WHERE
    b.g_idcard = '440401430103082'

-- 4、删除g_ _cardapplydetail 表中所有姓李的记

DELETE 
FROM
  g_cardapplydetail 
WHERE
  g_applyno = ( SELECT g_applyno FROM ( SELECT g_applyno FROM g_cardapplydetail WHERE g_name LIKE '李%' ) aa );

java的字符类型采用的编码方案是Unicode

构造函数:方法名与类名相同   主要作用是完成对垒的初始化工作 一般在创建新对象时，系统自动调用构造函数。

设三个变量x=1,y=2,z=3,则表达式y+=z--/++x的值为？

i++ 是先引用后增加 ,先在i所在的表达式中使用i的当前值，后让i加1

++i 是先增加后引用,让i先加1，然后在i所在的表达式中使用i的新值

他们其实都是i=i+1的意思，但是在程序中运行的时候的执行的顺序不一样。

i-- 和--i  也同理

上题y=2+(z--/++x)=2+(3/2)=3

字符串压缩。

利用字符重复出现的次数，编写一种方法，实现基本的字符串压缩功能。比如，字符串aabcccccaaa会变为a2b1c5a3。若“压缩”后的字符串没有变短，则返回原先的字符串。你可以假设字符串中只包含大小写英文字母（a至z）。

输入："aabcccccaaa"

输出："a2b1c5a3"

输入："abbccd"

输出："abbccd"

解释："abbccd"压缩后为"a1b2c2d1"，比原字符串长度更长。

class Solution {
    public String compressString(String S) {
        if (S.length() == 0) { // 空串处理
            return S;
        }
        StringBuffer ans = new StringBuffer();
        int cnt = 1;
        char ch = S.charAt(0);
        for (int i = 1; i < S.length(); ++i) {
            if (ch == S.charAt(i)) {
                cnt++;
            } else {
                ans.append(ch);
                ans.append(cnt);
                ch = S.charAt(i);
                cnt = 1;
            }
        }
        ans.append(ch);
        ans.append(cnt);
        return ans.length() >= S.length() ? S : ans.toString();
    }
}

基本表结构：

      student(sno,sname,sage,ssex)学生表
       course(cno,cname,tno) 课程表
       sc(sno,cno,score) 成绩表

     teacher(tno,tname) 教师表

101，查询课程1的成绩比课程2的成绩高的所有学生的学号
select a.sno from
(select sno,score from sc where cno=1) a,

(select sno,score from sc where cno=2) b
where a.score>b.score and a.sno=b.sno

102，查询平均成绩大于60分的同学的学号和平均成绩

select a.sno as "学号", avg(a.score) as "平均成绩"

from

(select sno,score from sc) a

group by sno having avg(a.score)>60

103，查询所有同学的学号、姓名、选课数、总成绩

select a.sno as 学号, b.sname as 姓名,

count(a.cno) as 选课数, sum(a.score) as 总成绩

from sc a, student b

where a.sno = b.sno
group by a.sno, b.sname

或者：

selectstudent.sno as 学号, student.sname as 姓名,

count(sc.cno) as 选课数, sum(score) as 总成绩

from student left Outer join sc on student.sno = sc.sno

group by student.sno, sname

104，查询姓“张”的老师的个数

selectcount(distinct(tname)) from teacher where tname like '张%‘

或者：

select tname as "姓名", count(distinct(tname)) as "人数"

from teacher

where tname like'张%'
group by tname

105，查询没学过“张三”老师课的同学的学号、姓名

select student.sno,student.sname from student

where sno not in (select distinct(sc.sno) from sc,course,teacher

where sc.cno=course.cno and teacher.tno=course.tno and teacher.tname='张三')

106，查询同时学过课程1和课程2的同学的学号、姓名
select sno, sname from student

where sno in (select sno from sc where sc.cno = 1)
and sno in (select sno from sc where sc.cno = 2)
或者：

selectc.sno, c.sname from

(select sno from sc where sc.cno = 1) a,

(select sno from sc where sc.cno = 2) b,

student c

where a.sno = b.sno and a.sno = c.sno

或者：

select student.sno,student.sname from student,sc where student.sno=sc.sno and sc.cno=1

and exists( select * from sc as sc_2 where sc_2.sno=sc.sno and sc_2.cno=2)

107，查询学过“李四”老师所教所有课程的所有同学的学号、姓名

select a.sno, a.sname from student a, sc b

where a.sno = b.sno and b.cno in

(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四')

或者：

select a.sno, a.sname from student a, sc b,

(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四') e

where a.sno = b.sno and b.cno = e.cno

108，查询课程编号1的成绩比课程编号2的成绩高的所有同学的学号、姓名

select a.sno, a.sname from student a,

(select sno, score from sc where cno = 1) b,

(select sno, score from sc where cno = 2) c

where b.score > c.score and b.sno = c.sno and a.sno = b.sno

109，查询所有课程成绩小于60分的同学的学号、姓名

select sno,sname from student

where sno not in (select distinct sno from sc where score > 60)

110，查询至少有一门课程与学号为1的同学所学课程相同的同学的学号和姓名

select distinct a.sno, a.sname

from student a, sc b

where a.sno <> 1 and a.sno=b.sno and
b.cno in (select cno from sc where sno = 1)

或者：

select s.sno,s.sname

from student s,

(select sc.sno

from sc

where sc.cno in (select sc1.cno from sc sc1 where sc1.sno=1)and sc.sno<>1

group by sc.sno)r1

where r1.sno=s.sno

基本表结构：

      student(sno,sname,sage,ssex)学生表
       course(cno,cname,tno) 课程表

       sc(sno,cno,score) 成绩表

       teacher(tno,tname) 教师表

111、把“sc”表中“王五”所教课的成绩都更改为此课程的平均成绩

update sc set score = (select avg(sc_2.score) from sc sc_2 wheresc_2.cno=sc.cno)

from course,teacher where course.cno=sc.cno and course.tno=teacher.tno andteacher.tname='王五'

112、查询和编号为2的同学学习的课程完全相同的其他同学学号和姓名

这一题分两步查：

1，

select sno
from sc
where sno <> 2

group by sno
having sum(cno) = (select sum(cno) from sc where sno = 2)

2，
select b.sno, b.sname

from sc a, student b

where b.sno <> 2 and a.sno = b.sno

group by b.sno, b.sname

having sum(cno) = (select sum(cno) from sc where sno = 2)

113、删除学习“王五”老师课的sc表记录

delete sc from course, teacher

where course.cno = sc.cno and course.tno = teacher.tno and tname = '王五'

114、向sc表中插入一些记录，这些记录要求符合以下条件：

将没有课程3成绩同学的该成绩补齐, 其成绩取所有学生的课程2的平均成绩

insert sc select sno, 3, (select avg(score) from sc where cno = 2)

from student

where sno not in (select sno from sc where cno = 3)

115、按平平均分从高到低显示所有学生的如下统计报表：

-- 学号,企业管理,马克思,UML,数据库,物理,课程数,平均分

select sno as 学号

,max(case when cno = 1 then score end) AS 企业管理

,max(case when cno = 2 then score end) AS 马克思

,max(case when cno = 3 then score end) AS UML

,max(case when cno = 4 then score end) AS 数据库

,max(case when cno = 5 then score end) AS 物理

,count(cno) AS 课程数

,avg(score) AS 平均分

FROM sc

GROUP by sno

ORDER by avg(score) DESC

116、查询各科成绩最高分和最低分：

以如下形式显示：课程号，最高分，最低分
select cno as 课程号, max(score) as 最高分, min(score) 最低分
from sc group by cno

select  course.cno as '课程号'

,MAX(score) as '最高分'

,MIN(score) as '最低分'

from sc,course

where sc.cno=course.cno

group by course.cno

117、按各科平均成绩从低到高和及格率的百分数从高到低顺序

SELECT t.cno AS 课程号,

max(course.cname)AS 课程名,

isnull(AVG(score),0) AS 平均成绩,

100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/count(1) AS 及格率

FROM sc t, course

where t.cno = course.cno

GROUP BY t.cno
ORDER BY 及格率 desc

118、查询如下课程平均成绩和及格率的百分数(用"1行"显示):

企业管理（001），马克思（002），UML （003），数据库（004）

select

avg(case when cno = 1 then score end) as 平均分1,

avg(case when cno = 2 then score end) as 平均分2,

avg(case when cno = 3 then score end) as 平均分3,

avg(case when cno = 4 then score end) as 平均分4,

100 * sum(case when cno = 1 and score > 60 then 1 else 0 end) / sum(casewhen cno = 1 then 1 else 0 end) as 及格率1,

100 * sum(case when cno = 2 and score > 60 then 1 else 0 end) / sum(casewhen cno = 2 then 1 else 0 end) as 及格率2,

100 * sum(case when cno = 3 and score > 60 then 1 else 0 end) / sum(casewhen cno = 3 then 1 else 0 end) as 及格率3,

100 * sum(case when cno = 4 and score > 60 then 1 else 0 end) / sum(casewhen cno = 4 then 1 else 0 end) as 及格率4

from sc

119、查询不同老师所教不同课程平均分, 从高到低显示

select max(c.tname) as 教师, max(b.cname) 课程, avg(a.score) 平均分

from sc a, course b, teacher c

where a.cno = b.cno and b.tno = c.tno

group by a.cno

order by 平均分 desc

或者：

select r.tname as '教师',r.rname as '课程' , AVG(score) as '平均分'

from sc,

(select  t.tname,c.cno as rcso,c.cname as rname

from teacher t ,course c

where t.tno=c.tno)r

where sc.cno=r.rcso

group by sc.cno,r.tname,r.rname

order by AVG(score) desc

120、查询如下课程成绩均在第3名到第6名之间的学生的成绩：

-- [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩

select top 6 max(a.sno) 学号, max(b.sname) 姓名,

max(case when cno = 1 then score end) as 企业管理,

max(case when cno = 2 then score end) as 马克思,

max(case when cno = 3 then score end) as UML,

max(case when cno = 4 then score end) as 数据库,

avg(score) as 平均分

from sc a, student b

where a.sno not in

(select top 2 sno from sc where cno = 1 order by score desc)

 and a.sno not in (select top 2 sno from sc where cno = 2 order by scoredesc)

 and a.sno not in (select top 2 sno from sc where cno = 3 order by scoredesc)

 and a.sno not in (select top 2 sno from sc where cno = 4 order by scoredesc)

 and a.sno = b.sno

group by a.sno