ITPub博客
首页 博客 专栏 学院 社区 IT新媒体评选
发博

首页 > IT职业 > IT职场 > 玩转数据结构之集合(Set)和映射(Map)

玩转数据结构之集合(Set)和映射(Map)

IT职场 作者：mouser 时间：2021-09-09 09:34:00 0 删除 编辑

集合

图片描述
图片描述

以二分搜索树为底层的集合实现

基于上节的二分搜索树

public class BSTSet> implements Set {

    private BST bst;

    public BSTSet(){
        bst = new BST();
    }

    @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);
    }
}

以链表为底层的集合实现

import java.util.ArrayList;

public class LinkedListSet implements Set {

    private LinkedList list;

    public LinkedListSet(){
        list = new LinkedList();
    }

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

            LinkedListSet set1 = new LinkedListSet();
            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 words2 = new ArrayList();
        if(FileOperation.readFile("a-tale-of-two-cities.txt", words2)){
            System.out.println("Total words: " + words2.size());

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

时间复杂度测试

当二叉树只有一条的时候，时间复杂度和链表的相同。

图片描述

import java.util.ArrayList;

public class Main {

    private static double testSet(Set set, String filename){

        long startTime = System.nanoTime();

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

        System.out.println();

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

    }
}

LeetCode 804 唯一摩尔斯密码词

// Leetcode 804. Unique Morse Code Words
// https://leetcode.com/problems/unique-morse-code-words/description/

import java.util.TreeSet;

public class Solution {

    public int uniqueMorseRepresentations(String[] words) {

        String[] codes = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        TreeSet set = new TreeSet();
        for(String word: words){
            StringBuilder res = new StringBuilder();
            for(int i = 0 ; i

映射（字典）Map

图片描述

以链表为底层的映射实现

import java.util.ArrayList;

public class LinkedListMap implements Map {

    private class Node{
        public K key;
        public V value;
        public Node next;

        public Node(K key, V value, Node next){
            this.key = key;
            this.value = value;
            this.next = next;
        }

        public Node(K key, V value){
            this(key, value, null);
        }

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

        @Override
        public String toString(){
            return key.toString() + " : " + value.toString();
        }
    }

    private Node dummyHead;
    private int size;

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

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

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

    private Node getNode(K key){
        Node cur = dummyHead.next;
        while(cur != null){
            if(cur.key.equals(key))
                return cur;
            // 循环
            cur = cur.next;
        }
        return null;
    }

    @Override
    public boolean contains(K key){
        return getNode(key) != null;
    }

    @Override
    public V get(K key){
        Node node = getNode(key);
        return node == null ? null : node.value;
    }

    @Override
    public void add(K key, V value){
        Node node = getNode(key);
        if(node == null){
            dummyHead.next = new Node(key, value, dummyHead.next);
            size ++;
        }
        else
            node.value = value;
    }

    @Override
    public void set(K key, V newValue){
        Node node = getNode(key);
        if(node == null)
            throw new IllegalArgumentException(key + " doesn't exist!");

        node.value = newValue;
    }

    @Override
    public V remove(K key){

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

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

        return null;
    }

    public static void main(String[] args){

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

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

            LinkedListMap map = new LinkedListMap();
            for (String word : words) {
                if (map.contains(word))
                    // 频率加1   
                    map.set(word, map.get(word) + 1);
                else
                    map.add(word, 1);
            }

            System.out.println("Total different words: " + map.getSize());
            System.out.println("Frequency of PRIDE: " + map.get("pride"));
            System.out.println("Frequency of PREJUDICE: " + map.get("prejudice"));
        }

        System.out.println();
    }
}

以二分搜索树为底层的映射实现

import java.util.ArrayList;

public class BSTMap, V> implements Map {

    private class Node{
        public K key;
        public V value;
        public Node left, right;

        public Node(K key, V value){
            this.key = key;
            this.value = value;
            left = null;
            right = null;
        }
    }

    private Node root;
    private int size;

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

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

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

    // 向二分搜索树中添加新的元素(key, value)
    @Override
    public void add(K key, V value){
        root = add(root, key, value);
    }

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

        if(node == null){
            size ++;
            return new Node(key, value);
        }

        if(key.compareTo(node.key)  0)
            node.right = add(node.right, key, value);
        else // key.compareTo(node.key) == 0
            node.value = value;

        return node;
    }

    // 返回以node为根节点的二分搜索树中，key所在的节点
    private Node getNode(Node node, K key){

        if(node == null)
            return null;

        if(key.equals(node.key))
            return node;
        else if(key.compareTo(node.key)  0)
            return getNode(node.right, key);
    }

    @Override
    public boolean contains(K key){
        return getNode(root, key) != null;
    }

    @Override
    public V get(K key){

        Node node = getNode(root, key);
        return node == null ? null : node.value;
    }

    @Override
    public void set(K key, V newValue){
        Node node = getNode(root, key);
        if(node == null)
            throw new IllegalArgumentException(key + " doesn't exist!");

        node.value = newValue;
    }

    // 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);
    }

    // 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键为key的节点
    @Override
    public V remove(K key){

        Node node = getNode(root, key);
        if(node != null){
            root = remove(root, key);
            return node.value;
        }
        return null;
    }

    private Node remove(Node node, K key){

        if( node == null )
            return null;

        if( key.compareTo(node.key)  0 ){
            node.right = remove(node.right, key);
            return node;
        }
        else{   // key.compareTo(node.key) == 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;
        }
    }

    public static void main(String[] args){

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

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

            BSTMap map = new BSTMap();
            for (String word : words) {
                if (map.contains(word))
                    map.set(word, map.get(word) + 1);
                else
                    map.add(word, 1);
            }

            System.out.println("Total different words: " + map.getSize());
            System.out.println("Frequency of PRIDE: " + map.get("pride"));
            System.out.println("Frequency of PREJUDICE: " + map.get("prejudice"));
        }

        System.out.println();
    }
}

Test

import java.util.ArrayList;

public class Main {

    public static void main(String[] args) {

        ArrayList words = new ArrayList();
        if(FileOperation.readFile("pride-and-prejudice.txt", words) &&
                FileOperation.readFile("a-tale-of-two-cities.txt", words)){

            // Test BST
            long startTime = System.nanoTime();

            BSTSet set = new BSTSet();
            for(String word: words)
                set.add(word);

            for(String word: words)
                set.contains(word);

            long endTime = System.nanoTime();

            double time = (endTime - startTime) / 1000000000.0;

            System.out.println("Total different words: " + set.getSize());
            System.out.println("BSTSet: " + time + " s");

            // ---

            // Test TreeMap Trie
            startTime = System.nanoTime();

            Trie trie = new Trie();
            for(String word: words)
                trie.add(word);

            for(String word: words)
                trie.contains(word);

            endTime = System.nanoTime();

            time = (endTime - startTime) / 1000000000.0;

            System.out.println("Total different words: " + trie.getSize());
            System.out.println("TreeMap Trie: " + time + " s");

            // ---

            // Test HashMap Trie
            startTime = System.nanoTime();

            Trie2 trie2 = new Trie2();
            for(String word: words)
                trie2.add(word);

            for(String word: words)
                trie2.contains(word);

            endTime = System.nanoTime();

            time = (endTime - startTime) / 1000000000.0;

            System.out.println("Total different words: " + trie.getSize());
            System.out.println("HashMap Trie: " + time + " s");

            // ---

            // Test Array(Map) Trie
            startTime = System.nanoTime();

            Trie3 trie3 = new Trie3();
            for(String word: words)
                trie3.add(word);

            for(String word: words)
                trie3.contains(word);

            endTime = System.nanoTime();

            time = (endTime - startTime) / 1000000000.0;

            System.out.println("Total different words: " + trie.getSize());
            System.out.println("Array(Map) Trie: " + time + " s");
        }
    }
}

Leetcode 349 两个数组的交集 使用集合

图片描述

Leetcode 350 两个数组的交集 使用映射

图片描述

来自 “ ITPUB博客 ” ，链接：https://blog.itpub.net/3173/viewspace-2804761/ ，如需转载，请注明出处，否则将追究法律责任。

0
0
上一篇： 浅谈NodeJs的模块机制
下一篇： php不常用而又实用的函数
请登录后发表评论 登录
全部评论
mouser

注册时间：2019-08-30

  • 博文量
    84
  • 访问量
    627

最新文章

支持我们 作者招募 用户协议 FAQ Contact Us

北京盛拓优讯信息技术有限公司. 版权所有  京ICP备16024965号-8 北京市公安局海淀分局网监中心备案编号：11010802021510 niuxiaotong@pcpop.com 17352615567 未成年人举报专区

广播电视节目制作经营许可证(京) 字第1234号 中国互联网协会会员

违法和不良信息举报电话:18101376593