集合Set

集合的一个关键的特点就是不能存放重复的元素，二分搜索树是一个非常好的实现集合的底层数据结构

1、二分搜索树实现集合：

set接口

package Set;

public interface Set<E> {

    void add(E e);

    boolean contains(E e);

    void remove(E e);

    int getSize();

    boolean isEmpty();
}

BST.class

package Set;

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class BST<E extends Comparable<E>> {

    private class Node{
        public E e;
        public Node left, right;

        public Node(E e){
            this.e = e;
            left = null;
            right = null;
        }
    }

    private Node root;
    private int size;

    public BST(){
        root = null;
        size = 0;
    }

    public int size(){
        return size;
    }

    public boolean isEmpty(){
        return size == 0;
    }

    // 向二分搜索树中添加新的元素e
    public void add(E e){
        root = add(root, e);
    }

    // 向以node为根的二分搜索树中插入元素e，递归算法
    // 返回插入新节点后二分搜索树的根
    private Node add(Node node, E e){

        if(node == null){
            size ++;
            return new Node(e);
        }

        if(e.compareTo(node.e) < 0)
            node.left = add(node.left, e);
        else if(e.compareTo(node.e) > 0)
            node.right = add(node.right, e);

        return node;
    }

    // 看二分搜索树中是否包含元素e
    public boolean contains(E e){
        return contains(root, e);
    }

    // 看以node为根的二分搜索树中是否包含元素e, 递归算法
    private boolean contains(Node node, E e){

        if(node == null)
            return false;

        if(e.compareTo(node.e) == 0)
            return true;
        else if(e.compareTo(node.e) < 0)
            return contains(node.left, e);
        else // e.compareTo(node.e) > 0
            return contains(node.right, e);
    }

    // 二分搜索树的前序遍历
    public void preOrder(){
        preOrder(root);
    }

    // 前序遍历以node为根的二分搜索树, 递归算法
    private void preOrder(Node node){

        if(node == null)
            return;

        System.out.println(node.e);
        preOrder(node.left);
        preOrder(node.right);
    }

    // 二分搜索树的非递归前序遍历
    public void preOrderNR(){

        Stack<Node> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()){
            Node cur = stack.pop();
            System.out.println(cur.e);

            if(cur.right != null)
                stack.push(cur.right);
            if(cur.left != null)
                stack.push(cur.left);
        }
    }

    // 二分搜索树的中序遍历
    public void inOrder(){
        inOrder(root);
    }

    // 中序遍历以node为根的二分搜索树, 递归算法
    private void inOrder(Node node){

        if(node == null)
            return;

        inOrder(node.left);
        System.out.println(node.e);
        inOrder(node.right);
    }

    // 二分搜索树的后序遍历
    public void postOrder(){
        postOrder(root);
    }

    // 后序遍历以node为根的二分搜索树, 递归算法
    private void postOrder(Node node){

        if(node == null)
            return;

        postOrder(node.left);
        postOrder(node.right);
        System.out.println(node.e);
    }

    // 二分搜索树的层序遍历
    public void levelOrder(){

        Queue<Node> q = new LinkedList<>();
        q.add(root);
        while(!q.isEmpty()){
            Node cur = q.remove();
            System.out.println(cur.e);

            if(cur.left != null)
                q.add(cur.left);
            if(cur.right != null)
                q.add(cur.right);
        }
    }

    // 寻找二分搜索树的最小元素
    public E minimum(){
        if(size == 0)
            throw new IllegalArgumentException("BST is empty!");

        return minimum(root).e;
    }

    // 返回以node为根的二分搜索树的最小值所在的节点
    private Node minimum(Node node){
        if(node.left == null)
            return node;
        return minimum(node.left);
    }

    // 寻找二分搜索树的最大元素
    public E maximum(){
        if(size == 0)
            throw new IllegalArgumentException("BST is empty");

        return maximum(root).e;
    }

    // 返回以node为根的二分搜索树的最大值所在的节点
    private Node maximum(Node node){
        if(node.right == null)
            return node;

        return maximum(node.right);
    }

    // 从二分搜索树中删除最小值所在节点, 返回最小值
    public E removeMin(){
        E ret = minimum();
        root = removeMin(root);
        return ret;
    }

    // 删除掉以node为根的二分搜索树中的最小节点
    // 返回删除节点后新的二分搜索树的根
    private Node removeMin(Node node){

        if(node.left == null){
            Node rightNode = node.right;
            node.right = null;
            size --;
            return rightNode;
        }

        node.left = removeMin(node.left);
        return node;
    }

    // 从二分搜索树中删除最大值所在节点
    public E removeMax(){
        E ret = maximum();
        root = removeMax(root);
        return ret;
    }

    // 删除掉以node为根的二分搜索树中的最大节点
    // 返回删除节点后新的二分搜索树的根
    private Node removeMax(Node node){

        if(node.right == null){
            Node leftNode = node.left;
            node.left = null;
            size --;
            return leftNode;
        }

        node.right = removeMax(node.right);
        return node;
    }

    // 从二分搜索树中删除元素为e的节点
    public void remove(E e){
        root = remove(root, e);
    }

    // 删除掉以node为根的二分搜索树中值为e的节点, 递归算法
    // 返回删除节点后新的二分搜索树的根
    private Node remove(Node node, E e){

        if( node == null )
            return null;

        if( e.compareTo(node.e) < 0 ){
            node.left = remove(node.left , e);
            return node;
        }
        else if(e.compareTo(node.e) > 0 ){
            node.right = remove(node.right, e);
            return node;
        }
        else{   // e.compareTo(node.e) == 0

            // 待删除节点左子树为空的情况
            if(node.left == null){
                Node rightNode = node.right;
                node.right = null;
                size --;
                return rightNode;
            }

            // 待删除节点右子树为空的情况
            if(node.right == null){
                Node leftNode = node.left;
                node.left = null;
                size --;
                return leftNode;
            }

            // 待删除节点左右子树均不为空的情况

            // 找到比待删除节点大的最小节点, 即待删除节点右子树的最小节点
            // 用这个节点顶替待删除节点的位置
            Node successor = minimum(node.right);
            successor.right = removeMin(node.right);
            successor.left = node.left;

            node.left = node.right = null;

            return successor;
        }
    }

    @Override
    public String toString(){
        StringBuilder res = new StringBuilder();
        generateBSTString(root, 0, res);
        return res.toString();
    }

    // 生成以node为根节点，深度为depth的描述二叉树的字符串
    private void generateBSTString(Node node, int depth, StringBuilder res){

        if(node == null){
            res.append(generateDepthString(depth) + "null\n");
            return;
        }

        res.append(generateDepthString(depth) + node.e +"\n");
        generateBSTString(node.left, depth + 1, res);
        generateBSTString(node.right, depth + 1, res);
    }

    private String generateDepthString(int depth){
        StringBuilder res = new StringBuilder();
        for(int i = 0 ; i < depth ; i ++)
            res.append("--");
        return res.toString();
    }
}

BSTSet.class

package Set;

public class BSTSet<E extends Comparable<E>> implements Set<E> {

    private BST<E> bst;

    public BSTSet() {
        bst = new BST<E>();
    }

    @Override
    public int getSize() {
        return bst.size();
    }

    @Override
    public boolean isEmpty() {
        return bst.isEmpty();
    }

    @Override
    public void add(E e) {
        bst.add(e);
    }

    @Override
    public boolean contains(E e) {
        return bst.contains(e);
    }

    @Override
    public void remove(E e) {
        bst.remove(e);
    }
}

例：词汇量统计 相同的词汇量只统计一次

实现原理：

我们先读取《傲慢与偏见》这本书的所有单词，然后将它存放在一个数组里，这个数组里面的单词是包括重复的，然后我们再建一个以二分搜索树实现的集合Set,遍历数组，将里面的所有单词再放在集合set中，因为二分搜索树不允许包括重复的元素，所以这时就可以实现相同的统计量只统计一次了。

FileOperation.class

package Set;

import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Locale;
import java.util.Scanner;

// 文件相关操作
public class FileOperation {

    // 读取文件名称为filename中的内容，并将其中包含的所有词语放进words中
    public static boolean readFile(String filename, ArrayList<String> words) {

        if (filename == null || words == null) {
            System.out.println("filename is null or words is null");
            return false;
        }

        // 文件读取
        Scanner scanner;

        try {
            File file = new File(filename);
            if (file.exists()) {
                FileInputStream fis = new FileInputStream(file);
                scanner = new Scanner(new BufferedInputStream(fis), "UTF-8");
                scanner.useLocale(Locale.ENGLISH);
            } else
                return false;
        } catch (IOException ioe) {
            System.out.println("Cannot open " + filename);
            return false;
        }

        // 简单分词
        // 这个分词方式相对简陋, 没有考虑很多文本处理中的特殊问题
        // 在这里只做demo展示用
        if (scanner.hasNextLine()) {

            String contents = scanner.useDelimiter("\\A").next();

            int start = firstCharacterIndex(contents, 0);
            for (int i = start + 1; i <= contents.length(); )
                if (i == contents.length() || !Character.isLetter(contents.charAt(i))) {
                    String word = contents.substring(start, i).toLowerCase();
                    words.add(word);
                    start = firstCharacterIndex(contents, i);
                    i = start + 1;
                } else
                    i++;
        }

        return true;
    }

    // 寻找字符串s中，从start的位置开始的第一个字母字符的位置
    private static int firstCharacterIndex(String s, int start) {

        for (int i = start; i < s.length(); i++)
            if (Character.isLetter(s.charAt(i)))
                return i;
        return s.length();
    }
}

测试Main.class

package Set;

import java.util.ArrayList;

public class Main {

    public static void main(String[] args) {

        System.out.println("Pride and Prejudice");

        ArrayList<String> words1 = new ArrayList<>();
        if (FileOperation.readFile("pride-and-prejudice.txt", words1)) {
            System.out.println("Total words 包含重复的: " + words1.size());
            BSTSet<String> set1 = new BSTSet<String>();
            for (String word : words1) {
                set1.add(word);
            }
            System.out.println("Total different words: " + set1.getSize());
        }

        System.out.println();

        System.out.println("A Tale of Two Cities");

        ArrayList<String> words2 = new ArrayList<>();
        if (FileOperation.readFile("a-tale-of-two-cities.txt", words2)) {
            System.out.println("Total words: " + words2.size());

            BSTSet<String> set2 = new BSTSet<String>();
            for (String word : words2)
                set2.add(word);
            System.out.println("Total different words: " + set2.getSize());
        }
    }
}

测试结果：

2、链表实现集合：

LinkedList.class

package Set;

public class LinkedList<E> {

    private class Node{
        public E e;
        public Node next;

        public Node(E e, Node next){
            this.e = e;
            this.next = next;
        }

        public Node(E e){
            this(e, null);
        }

        public Node(){
            this(null, null);
        }

        @Override
        public String toString(){
            return e.toString();
        }
    }

    private Node dummyHead;
    private int size;

    public LinkedList(){
        dummyHead = new Node();
        size = 0;
    }

    // 获取链表中的元素个数
    public int getSize(){
        return size;
    }

    // 返回链表是否为空
    public boolean isEmpty(){
        return size == 0;
    }

    // 在链表的index(0-based)位置添加新的元素e
    // 在链表中不是一个常用的操作，练习用：）
    public void add(int index, E e){

        if(index < 0 || index > size)
            throw new IllegalArgumentException("Add failed. Illegal index.");

        Node prev = dummyHead;
        for(int i = 0 ; i < index ; i ++)
            prev = prev.next;

        prev.next = new Node(e, prev.next);
        size ++;
    }

    // 在链表头添加新的元素e
    public void addFirst(E e){
        add(0, e);
    }

    // 在链表末尾添加新的元素e
    public void addLast(E e){
        add(size, e);
    }

    // 获得链表的第index(0-based)个位置的元素
    // 在链表中不是一个常用的操作，练习用：）
    public E get(int index){

        if(index < 0 || index >= size)
            throw new IllegalArgumentException("Get failed. Illegal index.");

        Node cur = dummyHead.next;
        for(int i = 0 ; i < index ; i ++)
            cur = cur.next;
        return cur.e;
    }

    // 获得链表的第一个元素
    public E getFirst(){
        return get(0);
    }

    // 获得链表的最后一个元素
    public E getLast(){
        return get(size - 1);
    }

    // 修改链表的第index(0-based)个位置的元素为e
    // 在链表中不是一个常用的操作，练习用：）
    public void set(int index, E e){
        if(index < 0 || index >= size)
            throw new IllegalArgumentException("Set failed. Illegal index.");

        Node cur = dummyHead.next;
        for(int i = 0 ; i < index ; i ++)
            cur = cur.next;
        cur.e = e;
    }

    // 查找链表中是否有元素e
    public boolean contains(E e){
        Node cur = dummyHead.next;
        while(cur != null){
            if(cur.e.equals(e))
                return true;
            cur = cur.next;
        }
        return false;
    }

    // 从链表中删除index(0-based)位置的元素, 返回删除的元素
    // 在链表中不是一个常用的操作，练习用：）
    public E remove(int index){
        if(index < 0 || index >= size)
            throw new IllegalArgumentException("Remove failed. Index is illegal.");

        Node prev = dummyHead;
        for(int i = 0 ; i < index ; i ++)
            prev = prev.next;

        Node retNode = prev.next;
        prev.next = retNode.next;
        retNode.next = null;
        size --;

        return retNode.e;
    }

    // 从链表中删除第一个元素, 返回删除的元素
    public E removeFirst(){
        return remove(0);
    }

    // 从链表中删除最后一个元素, 返回删除的元素
    public E removeLast(){
        return remove(size - 1);
    }

    // 从链表中删除元素e
    public void removeElement(E e){

        Node prev = dummyHead;
        while(prev.next != null){
            if(prev.next.e.equals(e))
                break;
            prev = prev.next;
        }

        if(prev.next != null){
            Node delNode = prev.next;
            prev.next = delNode.next;
            delNode.next = null;
            size --;
        }
    }

    @Override
    public String toString(){
        StringBuilder res = new StringBuilder();

        Node cur = dummyHead.next;
        while(cur != null){
            res.append(cur + "->");
            cur = cur.next;
        }
        res.append("NULL");

        return res.toString();
    }
}

LinkedListSet.class

package Set;

import java.util.ArrayList;

public class LinkedListSet<E> implements Set<E> {

    private LinkedList<E> list;

    public LinkedListSet() {
        list = new LinkedList<E>();
    }

    @Override
    public int getSize() {
        return list.getSize();
    }

    @Override
    public boolean isEmpty() {
        return list.isEmpty();
    }

    @Override
    public void add(E e) {
        if (!list.contains(e))
            list.addFirst(e);
    }

    @Override
    public boolean contains(E e) {
        return list.contains(e);
    }

    @Override
    public void remove(E e) {
        list.removeElement(e);
    }

    public static void main(String[] args) {

        System.out.println("Pride and Prejudice");

        ArrayList<String> words1 = new ArrayList<>();
        if (FileOperation.readFile("pride-and-prejudice.txt", words1)) {
            System.out.println("Total words: " + words1.size());

            LinkedListSet<String> set1 = new LinkedListSet<String>();
            for (String word : words1)
                set1.add(word);
            System.out.println("Total different words: " + set1.getSize());
        }

        System.out.println();


        System.out.println("A Tale of Two Cities");

        ArrayList<String> words2 = new ArrayList<>();
        if (FileOperation.readFile("a-tale-of-two-cities.txt", words2)) {
            System.out.println("Total words: " + words2.size());

            LinkedListSet<String> set2 = new LinkedListSet<String>();
            for (String word : words2)
                set2.add(word);
            System.out.println("Total different words: " + set2.getSize());
        }
    }
}

3、两种实现方式时间复杂度的对比

package Set;

import java.util.ArrayList;

//时间复杂度对比
public class TimeComplexityofSet {
    private static double testSet(Set<String> set, String filename){

        long startTime = System.nanoTime();

        System.out.println(filename);
        ArrayList<String> words = new ArrayList<>();
        if(FileOperation.readFile(filename, words)) {
            System.out.println("Total words: " + words.size());

            for (String word : words)
                set.add(word);
            System.out.println("Total different words: " + set.getSize());
        }
        long endTime = System.nanoTime();

        return (endTime - startTime) / 1000000000.0;
    }

    public static void main(String[] args) {

        String filename = "pride-and-prejudice.txt";

        BSTSet<String> bstSet = new BSTSet<String>();
        double time1 = testSet(bstSet, filename);
        System.out.println("BST Set: " + time1 + " s");

        System.out.println();

        LinkedListSet<String> linkedListSet = new LinkedListSet<String>();
        double time2 = testSet(linkedListSet, filename);
        System.out.println("Linked List Set: " + time2 + " s");

    }
}

通过测试可以看出二分搜索树是高效的

二分搜索树的最坏情况，这样的二分搜索树和链表是无异的。换句话来说，二分搜索树可能退化成链表。如下：

来源：CSDN

作者：我_是_太_阳

链接：https://blog.csdn.net/qq_43229543/article/details/104309252