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数据结构-集合和映射

数据结构-集合和映射

作者: 听你讲故事啊 | 来源:发表于2019-04-01 21:00 被阅读0次

    Set

    不能存放重复元素
    接口方法

    public interface Set<E> {

    void add(E e);
    boolean contains(E e);
    void remove(E e);
    int getSize();
    boolean isEmpty();
}

    二分搜索树实现

    借助前面的二分搜索树,可以很轻松的实现Set

    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);
    }
}

    链表实现

    使用前面创建的LinkedListSet来实现

    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);
    }

}

    时间复杂度分析

    编写一个main函数进行测试

    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");

    }
}

    结果

    pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
BST Set: 0.1498369 s

pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
Linked List Set: 3.2225657 s
    LinkedListSet BSTSet
    增 add O(n) O(h)
    查 contains O(n) O(h)
    删 remove O(n) O(h)

    h表示树的深度, h和n的关系如下:
    假设树为满二叉树, 根节点所在的那一层算是第0层

    层数 节点数
    0层 1
    1层 2
    2层 4
    3层 8
    4层 16
    h-1层 2^(h-1)

    这样算下来的话, h层, 共有 2^h-1个结点,
    2^h -1 = n ==> h=log₂(n+1)=O(log₂n)=O(logn)

    这样算下来,时间复杂度应该是:

    LinkedListSet BSTSet 最优
    增 add O(n) O(h) O(logn)
    查 contains O(n) O(h) O(logn)
    删 remove O(n) O(h) O(logn)

    如果二叉树退化成链表,那么时间复杂度就和链表集合一样了

    LinkedListSet BSTSet - 最优 - 最坏
    增 add O(n) O(h) - O(logn) - O(n)
    查 contains O(n) O(h) - O(logn) - O(n)
    删 remove O(n) O(h) - O(logn) - O(n)

    Map

    存储(键, 值)数据对的数据结构, 根据键去寻找值, 可以很轻松的使用链表或者二分搜索树实现

    class Node{
    K key;
    V value;
    Node next;
}

class Node{
    K key;
    V value;
    Node left;
    Node right;
}

    映射的接口类

    public interface Map<K, V> {
    void add(K key, V value);
    V remove(K key);
    boolean contains(K key);
    V get(K key);
    void set(K key, V newValue);
    int getSize();
    boolean isEmpty();
}

    链表实现

    需要对节点内容进行修改

    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;
            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;
    }
}

    二分搜索树实现

    修改一下节点内容, 其他的都可以复用二分搜索树中的方法

    
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;
        }
    }
}

    时间复杂度分析

    编写一个main函数进行测试

    import java.util.ArrayList;

public class Main {

    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;
    }

    public static void main(String[] args) {

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

        BSTMap<String, Integer> bstMap = new BSTMap<>();
        double time1 = testMap(bstMap, filename);
        System.out.println("BST Map: " + time1 + " s");

        System.out.println();

        LinkedListMap<String, Integer> linkedListMap = new LinkedListMap<>();
        double time2 = testMap(linkedListMap, filename);
        System.out.println("Linked List Map: " + time2 + " s");

    }
}

    运行结果

    pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
Frequency of PRIDE: 53
Frequency of PREJUDICE: 11
BST Map: 0.1996936 s

pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
Frequency of PRIDE: 53
Frequency of PREJUDICE: 11
Linked List Map: 14.6253311 s

    时间复杂度和集合类似

    LinkedListSet BSTSet - 最优 - 最坏
    增 add O(n) O(h) - O(logn) - O(n)
    删 remove O(n) O(h) - O(logn) - O(n)
    改 set O(n) O(h) - O(logn) - O(n)
    查 get O(n) O(h) - O(logn) - O(n)
    查 contains O(n) O(h) - O(logn) - O(n)

    对比

    image

    完整代码

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