1、有关集合Set的代码

1.1、Set.java

public interface Set<E> {

    void add(E e);

    void remove(E e);

    boolean contains(E e);

    int getSize();

    boolean isEmpty();

}

1.2、BSTSet.java

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

    private BST<E> bst;

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

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

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

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

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

    @Override
    public boolean isEmpty() {
        return false;
    }

    public static void main(String[] args) {
        System.out.println("Pride and Prejudice");

        ArrayList<String> words1 = new ArrayList<>();
        FileOperation.readFile(
                "D:\\idea_ws\\java\\JavaTest1\\src\\com\\ph\\pride-and-prejudice.txt", words1);

//        FileOperation.readFile(
//                "..\\..\\..\\pride-and-prejudice.txt", words1);
        System.out.println("Total words: " + words1.size());

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

    }

}

1.3、LinkedListSet.java

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

    private LearnLinkedList<E> list;

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

    /**
     * LearnLinkedList 可以有重复的元素
     * Set 不可以有重复的元素
     * 所以这里要稍作一些逻辑的处理
     */
    @Override
    public void add(E e) {
        if (!list.contains(e))
            list.addFirst(e);
    }

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

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

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

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

    public static void main(String[] args) {
        System.out.println("Pride and Prejudice");

        ArrayList<String> words1 = new ArrayList<>();
        FileOperation.readFile(
                "D:\\idea_ws\\java\\JavaTest1\\src\\com\\ph\\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());

    }
}

1.4、SetMainTest.java

/**
 * 测试 BSTSet 和 LinkedListSet 的性能
 */
public class SetMainTest {

    public static void main(String[] args) {
        String fileName = "D:\\idea_ws\\java\\JavaTest1\\src\\com\\ph\\pride-and-prejudice.txt";

        double time1 = testSet(new BSTSet<>(), fileName);
        System.out.println("BSTSet: " + time1 + "s");

        System.out.println();

        double time2 = testSet(new LinkedListSet<>(), fileName);
        System.out.println("LinkedListSet: " + time2 + "s");
    }

    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;
    }
}

打印结果：

可以看到BST二分搜索树的性能要远高于LinkedListSet。

下面我们进行时间复杂度分析。

2、集合的时间复杂度分析

其中的h是指二分搜索树的高度，那么h和n的关系是怎么样的呢？

如下：

分析1：

分析2：

分析3：

分析4：

假设数据量为100万，如果一个O（logn）算法需要一天的话，O（n）算法需要137年。

分析5：

3、Map 映射

称映射不是那么好理解，称为字典就更好理解一点。在Python中，map这种数据结构的名字就是dict（dictionary的缩写字典）、

总结：

存储（键，值）数据对的数据结构（Key，Value）
根据Key，寻找Value
非常容易使用链表或者二分搜索树实现

3.1、Map.java

public interface Map<K, V> {

    int getSize();

    boolean isEmpty();

    void add(K key, V value);

    V remove(K key);

    boolean contains(K key);

    V get(K key);

    void set(K key, V newValue);

}

3.2、LinkedListMap.java

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) {
            this(key, null, 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;
            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;
    }

    /**
     * 不能有两个相同的 key
     */
    @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 + "不存在！");

        node.value = newValue;
    }

    // 最复杂
    @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> words1 = new ArrayList<>();
        if (FileOperation.readFile(Constant.FILE_NAME, words1)) {
            System.out.println("Total words: " + words1.size());

            // 用来做单词频率统计，string用于放单词，integer用于放频率
            LinkedListMap<String, Integer> map = new LinkedListMap<>();

            for (String word : words1) {
                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"));
        }



    }

}

3.3、BSTMap.java

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;


    @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;
    }

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

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

    // 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(Constant.FILE_NAME, 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();
    }


}

3.4、MapMainTest.java

public class MapMainTest {

    public static void main(String[] args) {
        double time1 = testMap(new BSTMap<>(), Constant.FILE_NAME);
        System.out.println("BSTMap: " + time1 + "s");

        System.out.println();

        double time2 = testMap(new LinkedListMap<>(), Constant.FILE_NAME);
        System.out.println("LinkedListMap: " + time2 + "s");
    }

    private static double testMap(Map<String,Integer> map, 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) {
                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"));
        }

        long endTime = System.nanoTime();

        return (endTime - startTime) / 1000000000.0;
    }


}