集合 02 基于链表的集合-优快云博客

本文介绍了一种基于链表实现的集合类LinkedListSet，用于处理不重复元素的集合操作，通过词频分析测试其性能。对比BSTSet，虽然在速度上略显逊色，但在元素去重方面表现良好。

基于链表的集合

由于Set中的元素要求不能重复，故在添加元素时做了去重判断，这点相对于BSTSet要特别注意一下；

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);
    }

}

测试代码 - 词频分析

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());
    }
}

输出：

结果上和BSTSet是一样的，但明显可以感觉出速度上慢很多；
对于BSTSet和LinkedListSet的时间复杂度的分析，参见下一篇文章；

Pride and Prejudice
Total words: 125901
Total different words: 6530

A Tale of Two Cities
Total words: 141489
Total different words: 9944