【数据结构】—— 5、集合和映射_思考序列,集合,和映射在数据关系层面-优快云博客

本文链接：https://blog.youkuaiyun.com/mqy1023/article/details/94590999

之前用二分搜索树实现了集合
现在使用链表实现集合LinkedList

基于二分搜索树的集合实现

	class Node {
		E e;
		Node left;
		Node right;
	}

基于LinkedList链表的集合实现

	class Node {
		E e;
		Node next;
	}

映射

存储（键 key，值 value）数据对的数据结构
根据键（key），寻找值（value）
- 身份证号码----------------> 人
- 车牌号----------------------> 车
- 数据库 ID------------------> 信息
- 词频统计单词------------> 数字

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();
}

LinkedListMap 基于链表的映射实现

package map;

public class LinkedListMap<K, V> implements Map<K, V> {
    private class Node {
        private K key;
        private V value;
        public Node next;
        public Node(K key, V value, Node next) {
            key = key;
            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;
    }
    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 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 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;
    }
    @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 set(K key, V newValue) {
        Node node = getNode(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;
    }
}

BSTMap 基于二分搜索是树的映射实现

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;
    }

    // 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 void add(K key, V value) {

    }

    @Override
    public V remove(K key) {
        return null;
    }

    @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;
        }
    }
}