第一题：集合的运用（幸存者）
public class demo1 {public static void main(String[] args) {ArrayList<Integer> array = new ArrayList<>(); //一百个囚犯存放在array集合中Random r = new Random();for (int i = 0; i < 100; i++) {OUT:while (true) {int n = r.nextInt(200)+1; //生成随机数1-200//用for循环对比有没有重复比较复杂
// for (int j = 0; j < i; j++) {
// if(n==array.get(j)){ //随机数重复
// continue OUT;
// }
// }if(array.contains(n)){ //随机数重复continue OUT;}//说明没有重复的array.add(n);break OUT;}}System.out.println(array);ArrayList<Integer> arr = new ArrayList<>(); //保存一百个囚犯第一次的存放位置arr.addAll(array);System.out.println(arr);//题中要求位置从1开始计数，现在我们的位置是从0开始，所以后续计算要+1while(array.size()>1){for (int i = array.size()-1; i>=0; i--) {if((i+1)%2==1){ //奇数位置array.remove(i);}}System.out.println(array);}System.out.println("幸存者编号：" + array.get(0));System.out.print("幸存者第一次所占的位置（从1开始算）：");System.out.print(arr.indexOf(array.get(0)) + 1);}
}
第二题：基础编程能力
//User
public class User {private Long id; //用户idprivate String name; //用户名private String gender; //用户性别private LocalDate birthday; //用户生日public User() {}public User(Long id, String name, String gender, LocalDate birthday) {this.id = id;this.name = name;this.gender = gender;this.birthday = birthday;}public Long getId() {return id;}public void setId(Long id) {this.id = id;}public String getName() {return name;}public void setName(String name) {this.name = name;}public String getGender() {return gender;}public void setGender(String gender) {this.gender = gender;}public LocalDate getBirthday() {return birthday;}public void setBirthday(LocalDate birthday) {this.birthday = birthday;}@Overridepublic String toString() {return "User{" +"id=" + id +", name='" + name + '\'' +", gender='" + gender + '\'' +", birthday=" + birthday +'}';}
}//demo2
public class demo2 {public static void main(String[] args) {//创建一个ArrayList集合List<User> users = new ArrayList<>();//解析字符串String userStrs = "10001:张三:男:1990-01-01#10002:李四:女:1989-01-09#10003:王五:男:1999-09-09#10004:刘备:男:1899-01-01#10005:孙悟空:男:1900-01-01#10006:张三:女:1999-01-01#10007:刘备:女:1999-01-01#10008:张三:女:2003-07-01#10009:猪八戒:男:1900-01-01";String[] strs = userStrs.split("#"); //用#号把每个用户的数据先拆分开//System.out.println(Arrays.toString(strs)); //[10001:张三:男:1990-01-01, 10002:李四:女:1989-01-09, 10003:王五:男:1999-09-09, 10004:刘备:男:1899-01-01, 10005:孙悟空:男:1900-01-01, 10006:张三:女:1999-01-01, 10007:刘备:女:1999-01-01, 10008:张三:女:2003-07-01, 10009:猪八戒:男:1900-01-01]for (int i = 0; i < strs.length; i++) {String s = strs[i];String[] str2 = s.split(":"); //用:号把每个用户的个人数据（id 姓名 性别 生日）拆分开//str2[0]代表id str2[1]代表姓名 str2[2]代表性别 str2[3]代表生日long id = Long.parseLong(str2[0]); //把用户id从String型转换成long型//long id = Long.valueOf(str2[0]); //把用户id从String型转换成long型//把生日转换成从字符串转换成LocalDateLocalDate birth = LocalDate.parse(str2[3]);
// String[] time = str2[3].split("-"); //time[0]代表年 time[1]代表月 time[2]代表日
// int year = Integer.parseInt(time[0]);
// int month = Integer.parseInt(time[1]);
// int day = Integer.parseInt(time[2]);
// LocalDate birth = LocalDate.of(year,month,day);User user = new User(id,str2[1],str2[2],birth);users.add(user);}System.out.println(users);System.out.println("============================================");//遍历List<User>集合，统计每个名字出现的次数Map<String,Integer> map = new HashMap<>();for (int i = 0; i < users.size(); i++) {User user = users.get(i);if(map.containsKey(user.getName())){ //如果用户的名字在map的键中存在map.put(user.getName(),map.get(user.getName())+1);}else{ //如果用户的名字在map的键中第一次出现map.put(user.getName(),1);}}//System.out.println(map);map.forEach((k,v) -> System.out.println(k+":"+v+"次"));}
}
第三题：JDK8新时间的应用
public class demo3 {public static void main(String[] args) {LocalDate ld = LocalDate.of(2022,2,3); //记录首次休息日Scanner sc = new Scanner(System.in);while (true) {System.out.println("请输入您查询月份（XXXX-X）：");String s = sc.next();if(!s.matches("\\d{4}-\\d{1,2}")){System.out.println("您输入的时间格式不正确，请重新输入");continue;}else {String[] str = s.split("-");int year = Integer.parseInt(str[0]); //查询的年份int month = Integer.parseInt(str[1]); //查询的月份呢if(month > 12 || month < 1){ //查询的月份不是1-12之间System.out.println("请输入正确的月份（1-12）~~~");}else if (year < 2022 || (year == 22 && month <= 2)) { //查询月份不在2022年2月之后System.out.println("请输入2022年2月之后的月份~~~");} else { //查询月份在2022年2月之后 且 月份符合规范（1-12）int days = dayNum(year, month); //该月有几天//打印该月的上班情况for (int i = 1; i < days; i++) {LocalDate date = LocalDate.of(year, month, i);//判断该日是否是休息日Long next = date.toEpochDay() - ld.toEpochDay(); //获取相差天数if (next % 3 == 0) {System.out.print(date + "[休息]");//判断休息日是否是周末（周六 周日）if(date.getDayOfWeek() == DayOfWeek.SATURDAY){System.out.print("[周六] ");}else if(date.getDayOfWeek() == DayOfWeek.SUNDAY){System.out.println("[周日] ");}else{System.out.print(" ");}} else if (next % 3 == 1 || next % 3 == 2) {System.out.print(date + " ");}}break;}}}}public static int dayNum(int year, int month) {int num = 0;switch (month){case 1:case 3:case 5:case 7:case 8:case 10:case 12:num = 31;break;case 4:case 6:case 9:case 11:num = 30;break;case 2:if((year%4==0 && year%100!=0) || (year%400==0)){//闰年num = 29;}else{num = 28;}break;default:System.out.println("月份有误");break;}return num;}
}
第四题：手写ArrayList集合
//MyArrayList
public class MyArrayList<E> {private Object[] arr = new Object[10];private int count = 0; //记录数组中现存有几个数据private double gene = 0.8; //激活因子//往集合中添加数据（返回值是boolean类型）public boolean add(E e){arr[count] = e;count++;if(count >= arr.length * gene){Object[] arr1 = Arrays.copyOf(arr,arr.length * 2); //如果数组中的数据个数等于或超过数组最大范围的80%，则扩容两倍arr = arr1;}//System.out.println(Arrays.toString(arr));//但是由于用户只是当成集合，应该设计成看不到最后扩容的null值位//比如用户存入第一个数据11，用户希望返回的是[11]，不是[11,null,null…]
// Object[] rs = Arrays.copyOf(arr,count);
// System.out.println(Arrays.toString(rs));return true;}//根据索引查询指定元素public E get(int index) {//但是由于用户只是当成集合，应该设计成看不到最后扩容的null值位//比如用户存入第一个数据11，用户希望返回的是[11]，不是[11,null,null…]//因此用户输入超过他自己存入个数的索引值，就产生了越界if(index >= count || index < 0){ //索引值越界throw new ArrayOutException("您输入的索引越界");}else{return (E) arr[index];}}//根据索引删除指定元素（返回值是被删除的元素）public E remove(int index){if(index >= count || index < 0) { //索引值越界throw new ArrayOutException("您输入的索引越界");}else{E e = (E) arr[index]; //记录被删除元素//删除的元素是最后一个元素if(index == arr.length-1){ //由于扩容机制，所以不会删除的元素永远不会是数组的最后一个arr[index] = null;return e;}//删除的元素不是最后一个元素，需要进行移位（后面的数前移）for (int i = index + 1; i < arr.length; i++) {if(i != arr.length-1){arr[i-1] = arr[i];}else{arr[i] = null; //由于扩容机制，永远都没有存满，最后一位永远是null，因此最后一个数的前移不会导致最后一位多出来一个重复的数}}//System.out.println(Arrays.toString(arr));count--; //数组元素个数-1//但是由于用户只是当成集合，应该设计成看不到最后扩容的null值位//比如用户存入第一个数据11，用户希望返回的是[11]，不是[11,null,null…]
// Object[] rs = Arrays.copyOf(arr,count);
// System.out.println(Arrays.toString(rs));return e;}}//返回集合大小public int size(){return count;}//遍历集合public void forEach1(){for (int i = 0; i < count; i++) {System.out.print(arr[i]+ " ");}System.out.println();}//遍历集合（可以用Lambda表达式）public void forEach(MyConsumer<E> action){Objects.requireNonNull(action);for (int i = 0; i < count; i++) {action.accept((E) arr[i]);}}public String toString() {StringBuilder sb = new StringBuilder();sb.append("[");for (int i = 0; i < count; i++) {E e = (E) arr[i];sb.append(e).append(i==count-1?"":", "); //判断该元素是否是最后一个数据，是否需要加,}sb.append("]");return sb.toString();}
}//MyConsumer
public interface MyConsumer<E> {void accept(E e);
}//ArrayOutException
public class ArrayOutException extends RuntimeException{public ArrayOutException(){}public ArrayOutException(String message){super(message);}
}//demo4
public class demo4 {public static void main(String[] args) {MyArrayList<Integer> arr = new MyArrayList<>();arr.add(11);arr.add(22);arr.add(33);arr.add(44);arr.add(55);arr.add(66);arr.add(77);arr.add(88);arr.add(99);System.out.println(arr); //[11, 22, 33, 44, 55, 66, 77, 88, 99]System.out.println(arr.size()); //集合大小 9System.out.println(arr.get(8)); //索引8是99//System.out.println(arr.get(11)); //您输入的索引越界//System.out.println(arr.get(-1)); //您输入的索引越界System.out.println(arr.remove(7)); //返回被删除的元素88System.out.println(arr.size()); //集合大小 8arr.forEach1(); //11 22 33 44 55 66 77 99arr.forEach((Integer integer) -> System.out.print(integer + " ")); //11 22 33 44 55 66 77 99}
}
第五题：二分查找的应用
public class demo5 {public static void main(String[] args) {int[] nums = {};int target = 0;int[] rs = isExist(nums,target);System.out.println(Arrays.toString(rs));}//查找目标值对应的最左边的位置public static int getLeftIndex(int[] nums, int target){int rs = -1; //数据不存在为-1//二分查找int left = 0;int right = nums.length-1;while (left<=right){int middle = (left + right) / 2;if(nums[middle]==target){rs = middle; //先临时存放第一次找到目标值的位置//二分查找该元素的 左边 是否还存在目标值right = middle - 1;}else if(nums[middle] < target){left = middle + 1;}else if(nums[middle] > target){right = middle - 1;}}return rs;}//查找目标值对应的最右边的位置public static int getRightIndex(int[] nums, int target){int rs = -1; //数据不存在为-1//二分查找int left = 0;int right = nums.length-1;while (left<=right){int middle = (left + right) / 2;if(nums[middle]==target){rs = middle; //先临时存放第一次找到目标值的位置//二分查找该元素的 右边 是否还存在目标值left = middle + 1;}else if(nums[middle] < target){left = middle + 1;}else if(nums[middle] > target){right = middle - 1;}}return rs;}//复杂度O(log2n)public static int[] isExist(int[] nums, int target) {int[] rs = {-1,-1}; //记录返回值if(nums == null ||nums.length == 0){ //如果 数组不存在 或 数组为空return rs;}//数组不为空rs[0] = getLeftIndex(nums,target);rs[1] = getRightIndex(nums,target);return rs;}//复杂度O(n)public static int[] isExist1(int[] nums, int target) {int[] rs = new int[2];int count = 0; //记录第几次找到该数字for (int i = 0; i < nums.length; i++) {if(nums[i] == target && count == 0){ //第一次找到该数字rs[0] = i;count++;}else if(nums[i] == target && count != 0){ //不是第一次找到该数字rs[1] = i;count++;}}if(count == 0){ //说明未找到rs[0] = -1;rs[1] = -1;}return rs;}
}
第六题：手写链表、反转链表
//MyLinkedList
public class MyLinkedList<E> {private int size;/*** 定义了一个私有的内部类，作为链表的结点*/public static class Node<E>{E data;Node<E> next;public Node(E data, Node<E> next) {this.data = data;this.next = next;}}public Node<E> add(){Node<E> head = null;Scanner sc = new Scanner(System.in);while (true) {System.out.println("请您输入当前结点的数据值（exit为结束）：");String data = sc.next();if(data.equals("exit")){ //如果输入的是exit则结束break;}//输入不是exitif(head==null){ //第一次创建结点head = new Node(data,null);size++;}else{//已存在头结点，往后插入结点（尾插法）Node<E> temp = head;//让temp走到尾部while(temp.next != null){temp = temp.next;}//把当前结点值创建出来，加入尾部temp.next = new Node(data,null);size++;}}return head; //返回链表是返回链表的头结点}public Node<E> reverse(Node<E> head,int left,int right){if(head == null || left < 1 || left > size || right < 1 || right > size || left >= right){return head;}//反转//先找到左结点的位置//从左结点遍历到右结点，然后把数据存到集合中Node<E> first = head; //遍历结点标识Node<E> mark = null; //记录左结点List<E> data = new ArrayList<>();int index = 0;while(first != null){index++;if(index == left){mark = first;}if(index>=left && index<=right){data.add(first.data);}if(index == right){break;}first = first.next;}//倒序遍历集合for (int i = data.size()-1; i >= 0; i--) {E e = data.get(i);mark.data = e;mark = mark.next;}return head;}public void forEach(Node<E> head){if(head == null){System.out.println(head);return;}while(head != null){System.out.print(head.data+" ");head = head.next;}System.out.println();}}//demo6
public class demo6 {public static void main(String[] args) {MyLinkedList<String> list = new MyLinkedList<>();MyLinkedList.Node<String> head = list.add();list.forEach(head);MyLinkedList.Node<String> head2 = list.reverse(head,2,5);list.forEach(head2);}
}
