7-1 集合基础和基于二分搜索树的集合实现

public interface Set<E> {

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

利用上一章实现的二分搜索树

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);
    }
}

7-2 基于链表的集合实现

import java.util.ArrayList;

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);
    }
}

7-3 集合类的复杂度分析

7-4 LeetCode中的集合问题

LeetCode 804

import java.util.TreeSet;

public class Solution {

    public int uniqueMorseRepresentations(String[] words) {

        String[] codes = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        TreeSet<String> set = new TreeSet<>();
        for(String word: words){
            StringBuilder res = new StringBuilder();
            for(int i = 0 ; i < word.length() ; i ++)
                res.append(codes[word.charAt(i) - 'a']);

            set.add(res.toString());
        }

        return set.size();
    }
}

7-5 映射基础

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();
}

7-6 基于链表的映射实现

import java.util.ArrayList;

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;
    }
}

7-7 基于二分搜索树的映射实现

import java.util.ArrayList;

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;
        }
    }
}

7-8 映射的复杂度分析

7-9 LeetCode上更多集合和映射的问题

349. 两个数组的交集

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

class Solution349 {
    public int[] intersection(int[] nums1, int[] nums2) {

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

        ArrayList<Integer> list = new ArrayList<>();
        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;
    }
}

350. 两个数组的交集 II

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

public class Solution350 {

    public int[] intersect(int[] nums1, int[] nums2) {

        TreeMap<Integer, Integer> map = new TreeMap<>();
        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<>();
        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;
    }
}

最后

以上就是稳重河马为你收集整理的第7章 集合和映射7-1 集合基础和基于二分搜索树的集合实现7-2 基于链表的集合实现7-3 集合类的复杂度分析7-4 LeetCode中的集合问题7-5 映射基础7-6 基于链表的映射实现7-7 基于二分搜索树的映射实现7-8 映射的复杂度分析7-9 LeetCode上更多集合和映射的问题的全部内容，希望文章能够帮你解决第7章 集合和映射7-1 集合基础和基于二分搜索树的集合实现7-2 基于链表的集合实现7-3 集合类的复杂度分析7-4 LeetCode中的集合问题7-5 映射基础7-6 基于链表的映射实现7-7 基于二分搜索树的映射实现7-8 映射的复杂度分析7-9 LeetCode上更多集合和映射的问题所遇到的程序开发问题。

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