目录

映射map接口

 基于二分搜索树的map

 基于链表的map

 用集合求两个数组的交集

 用映射map求两个数组的交集

• 映射map接口

package com.suanfa.map;
/**
 * 映射Map接口
 * @author Administrator
 *
 */
public interface Map<K,V> {
	void add(K key,V value);//向map中添加键值对
	V remove(K key);//删除键值对
	boolean contains(K key);//判断是否存在某个键和值
	V get(K key);//通过键获取值
	void set(K key,V value);//修改键值对
	int getSize();//获取map大小
	boolean isEmpty();//判断map是否为空

}

•  基于二分搜索树的map

package com.suanfa.map;
/**
 * 基于二分搜索树的映射map
 * @author Administrator
 *
 */
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;
        }
    }

}

•  基于链表的map

package com.suanfa.map;
/**
 * 基于链表的映射map
 * @author Administrator
 *
 */
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;//虚拟头节点
    private int size;//元素个数

    public LinkedListMap(){
        dummyHead = new Node();
        size = 0;
    }

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

    @Override
    public boolean isEmpty(){
        return size == 0;
    }
    /**
     * 根据K获取节点
     * @param key
     * @return
     */
    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;
    }

}

•  用集合求两个数组的交集

package com.suanfa.map;

import java.util.ArrayList;
import java.util.TreeSet;

/**
 * 用集合求两个数组的交集
 * @author Administrator
 *
 */
public class Solution349 {
	public int[] intersection(int[] nums1, int[] nums2) {

        TreeSet<Integer> set = new TreeSet<Integer>();
        for(int num: nums1)
            set.add(num);

        ArrayList<Integer> list = new ArrayList<Integer>();
        for(int num: nums2){
            if(set.contains(num)){
                list.add(num);
                set.remove(num);
            }
        }

        int[] res = new int[list.size()];
        for(int i = 0 ; i < list.size() ; i ++)
            res[i] = list.get(i);
        return res;
    }

}

•  用映射map求两个数组的交集

package com.suanfa.map;

import java.util.ArrayList;
import java.util.TreeMap;

/**
 * 用映射map求两个数组的交集
 * @author Administrator
 *
 */
public class Solution350 {
	public int[] intersect(int[] nums1, int[] nums2) {

        TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
        for(int num: nums1){
            if(!map.containsKey(num))
                map.put(num, 1);
            else
                map.put(num, map.get(num) + 1);
        }

        ArrayList<Integer> res = new ArrayList<Integer>();
        for(int num: nums2){
            if(map.containsKey(num)){
                res.add(num);
                map.put(num, map.get(num) - 1);
                if(map.get(num) == 0)
                    map.remove(num);
            }
        }

        int[] ret = new int[res.size()];
        for(int i = 0 ; i < res.size() ; i ++)
            ret[i] = res.get(i);

        return ret;
    }

}