版权声明：本文为博主原创文章，转载请附上博文链接！ https://blog.csdn.net/qq_24095055/article/details/89052891
映射
映射概念的实现：
| dict | key | value |
|---|---|---|
| 字典 | 单词 | 释义 |
| 名册 | 身份证号 | 人 |
| 车辆管理 | 车牌号 | 车 |
| 数据库 | id | 信息 |
| 词频统计 | 单词 | 频率 |
- 存储（键，值）数据对的数据结构（Key，Value）
- 根据键（Key），寻找值（Value）
- 非常容易使用链表或者二分搜索树实现
Map接口定义
/**
* Created by binzhang on 2019/4/5.
*/
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();
}
基于链表的映射实现
/**
* Created by binzhang on 2019/4/5.
*/
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;
}
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;
}
}
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;
}
}
public static void main(String[] args){
System.out.println("Pride and Prejudice");
ArrayList<String> words = new ArrayList<>();
if(FileOperation.readFile("pride-and-prejudice.txt", 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();
}
}
映射的复杂度分析
有序映射和无序映射
- 有序映射中的键具有顺序性 基于搜索树的实现
- 无序映射中的键没有顺序性 基于哈希表的实现
多重映射
- 多重映射中的键可以重复