玩转数据结构之集合(Set)和映射(Map)

集合

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<E extends Comparable<E>> implements Set<E> {

    private BST<E> 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<E> implements Set<E> {

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

            LinkedListSet<String> 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<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<>();
            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。

import java.util.ArrayList;

public class Main {

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

        System.out.println();

        LinkedListSet<String> 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<String> set = new TreeSet<>();
        for(String word: words){
            StringBuilder res = new StringBuilder();
            for(int i = 0 ; i < word.length() ; i ++)
                res.append(codes[word.charAt(i) - 'a']);

            set.add(res.toString());
        }

        return set.size();
    }
}

映射（字典）Map

以链表为底层的映射实现

import java.util.ArrayList;

public class LinkedListMap<K, V> implements Map<K, V> {

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

            LinkedListMap<String, Integer> 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<K extends Comparable<K>, V> implements Map<K, V> {

    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.left = add(node.left, key, value);
        else 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.left, key);
        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.left = remove(node.left , key);
            return node;
        }
        else 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<String> words = new ArrayList<>();
        if(FileOperation.readFile("pride-and-prejudice.txt", words)) {
            System.out.println("Total words: " + words.size());

            BSTMap<String, Integer> 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<String> 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<String> 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

Leetcode 350 两个数组的交集 使用映射