第14章_集合与数据结构拓展练习

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选择填空题

1、前序、中序、后序遍历

分析：

完全二叉树： 叶结点只能出现在最底层的两层，且最底层叶结点均处于次底层叶结点的左侧

题1：
前序遍历：中左右  ABDECF中序遍历：左中右  DBEAFC后序遍历：左右中  DEBFCA

题2：n-i+1

2、线性结构

C

3、其它

1、先根次序遍历，就是前序遍历：
ABDHIECFG
2、队列先进先出
3、C
4、C
5、2的4次方是16个

编程题

4、单向链表构建

（1）定义一个单向链表SingleLinked类

• 包含私有的静态内部类Node

  • 包含Object类型的data属性和Node类型的next属性
  • 包含有参构造Node(Object data, Node next)

• 包含私有的单链表的Node类型的头结点first

• 包含public void add(Object element)方法，可以添加元素到当前单链表中

• 包含私有的非静态内部类Itr，Itr类实现java.util.Iterator接口

  • 包含Node类型的实例变量node，初始化为单链表的first
  • 重写boolean hasNext()方法，判断node结点是否为空
  • 重写Object next()方法，获取node对象的data值，并让node结点指向下一个结点

• 单向链表SingleLinked类实现java.lang.Iterable接口，

  • 重写Iterator iterator()方法，返回非静态内部类Itr的对象

（2）测试类中创建SingleLinked单链表的对象，并添加（张三、李四、王五、赵六）几个元素到单链表中，并用foreach循环变量输出。

public class SingleLinked implements Iterable{private Node first;//单向链表的头private static class Node{Object data;Node next;Node(Object data, Node node) {this.data = data;this.next = node;}}public void add(Object element){Node newNode = new Node(element, null);if(first == null){first = newNode;return;}Node node = first;while(node.next !=null){node = node.next;}node.next = newNode;}@Overridepublic Iterator iterator() {return new Itr();}private class Itr implements Iterator{Node node = first;@Overridepublic boolean hasNext() {return node != null;}@Overridepublic Object next() {Object element = node.data;node = node.next;return element;}}/* 暴露静态内部类public static class Knot{public Object data;public Knot next;}*/}

public class Exercise4 {public static void main(String[] args) {//违反了高内聚低耦合的原则
/*        SingleLinked.Knot k1 = new SingleLinked.Knot();k1.data = "张三";SingleLinked.Knot k2 = new SingleLinked.Knot();k2.data = "李四";k1.next = k2;*///高内聚低耦合SingleLinked link = new SingleLinked();link.add("张三");link.add("李四");link.add("王五");link.add("赵六");for (Object o : link) {System.out.println(o);}}
}

5、单向链表及其反转

单链表的实现

public class OneWayLinkedList<E>{private Node<E> head;private int total;private static class Node<E>{E data;Node<E> next;Node(E data, Node<E> next) {this.data = data;this.next = next;}}public void add(E e) {Node<E> newNode = new Node<>(e,null);if(head==null){head = newNode;}else{Node<E> node = head;while(node.next!=null){node = node.next;}node.next = newNode;}total++;}public void delete(E e) {Node<E> node = head;Node<E> find = null;Node<E> last = null;if(e==null){while(node!=null){if(node.data==null){find = node;break;}last = node;node = node.next;}}else{while(node!=null){if(e.equals(node.data)){find = node;break;}last = node;node = node.next;}}if(find != null){if(last==null){head = find.next;}else{last.next = find.next;}total--;}}public void update(E old, E value) {Node<E> node = head;Node<E> find = null;if(old==null){while(node!=null){if(node.data==null){find = node;break;}node = node.next;}}else{while(node!=null){if(old.equals(node.data)){find = node;break;}node = node.next;}}if(find != null){find.data = value;}}public boolean contains(E e) {return indexOf(e) != -1;}public int indexOf(E e) {int index = -1;if(e==null){int i=0;for(Node<E> node = head; node!=null; node=node.next ){if(node.data==e){index=i;break;}i++;}}else{int i=0;for(Node<E> node = head; node!=null; node=node.next ){if(e.equals(node.data)){index=i;break;}i++;}}return index;}public Object[] getAll() {Object[] all = new Object[total];Node<E> node = head;for (int i = 0; i < all.length; i++,node = node.next) {all[i] = node.data;}return all;}public int size() {return total;}@SuppressWarnings("unchecked")public void reverse() {if(head!=null) {Node<E>[] all = new Node[total];Node<E> node = head;for (int i = 0; i < all.length; i++) {all[i] = node;node = node.next;}head = all[all.length-1];node = head;for (int i = all.length-2; i >= 0; i--) {node.next = all[i];node = node.next;}}}
}

public class Exercise5 {public static void main(String[] args) {OneWayLinkedList<Integer> list = new OneWayLinkedList<>();for (int i = 1; i <= 5; i++) {list.add(i);}Object[] all = list.getAll();System.out.println(Arrays.toString(all));list.reverse();all = list.getAll();System.out.println(Arrays.toString(all));}
}

6、字符串压缩

实现简易字符串压缩算法，其中连续出现2次以上（含2次）的字母转换为字母和出现的次数。
例如：AAAABCCDEEEEE,压缩之后为A4BC2DE5。

代码实现：

public class Exercise6 {public static void main(String[] args) {
//		String str = "AAAABCCDEEEEE";String str = "AAAABCCDEEEEEF";LinkedList<String> list = new LinkedList<>();int count = 0;for (int i = 0; i < str.length(); i++) {if(list.size()==0) {list.addLast(str.charAt(i)+"");count++;}else {if(list.getLast().equals(str.charAt(i)+"")) {count++;}else {if(count>1) {list.addLast(count+"");}list.addLast(str.charAt(i)+"");count=1;}}}if(count>1) {list.addLast(count+"");}while(list.size()!=0) {System.out.print(list.pollFirst());}}
}