设计背景
在信息存储的场景中,我们经常遇到“一个值对应着另一个值”的情况,这种关系在数学中被称作函数(function)或者映射(mapping)。在数据结构中也有这样一种设计类型,每一个节点存储着一个key键和一个value值,两者合称“键值队”(key-value),用户可以通过key键定位到value值,这种结构类型被称为映射(Map)。
映射中key的值不能重复,而value得值可以重复,多个key可以映射到同一个value上。
结构分析
【底层实现】链表(LinkedList)/ 二分搜索树(BST)
【核心方法】
public void add(K key, V value); //向映射中添加键值队
public V remove(K key); //移除指定键值队
public void set(K key, V newValue); //修改指定键所对应的值
public V get(K key); //获取指定键所对应的值
【差异分析】
因为使用链表实现的映射(LinkedListMap)进行增加、删除等操作时,都需要对链表元素进行遍历,故其性能较低(时间复杂度为O(n)级别);而使用二分搜索树实现的映射(BSTMap)在理想状态下(满二叉树)最多只会遍历h个元素(时间复杂度为O(h)级别,从数值上看为O(logn)级别),所以通常BSTMap的性能与LinkedListMap的性能差异巨大。
代码实现
1. 利用LinkedList实现
public class LinkedListMap<K, V> implements Map<K, V> {
/**
* 内部类:节点
*/
private class Node {
/**
* 实例域:key值、value值、节点引用
*/
public K key;
public V value;
public Node next;
/**
* 构造器:对实例域进行初始化
*
* @param key 键
* @param value 值
* @param next 引用
*/
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);
}
}
private Node dummyHead;
private int size;
public LinkedListMap() {
dummyHead = new Node();
size = 0;
}
/**
* 方法:通过key值查找其所在的节点
*
* @param key 键
* @return 键所在的节点
*/
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 void set(K key, V newValue) {
Node node = getNode(key);
if (node == null) {
throw new IllegalArgumentException(key + "doesn't exists!");
}
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;
}
@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 int getSize() {
return size;
}
@Override
public boolean isEmpty() {
return false;
}
}
2. 利用BST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;
}
@Override
public void add(K key, V value) {
root = add(root, key, value);
}
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 {
node.value = value;
}
return node;
}
private Node getNode(Node node, K key) {
if (node == null) {
return null;
}
if (key.compareTo(node.key) == 0) {
return node;
} else if (key.compareTo(node.key) < 0) {
return getNode(node.left, key);
} else {
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 exists!");
}
node.value = newValue;
}
@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 {
if (node.left == null) {
Node rightNode = node.right;
node.right = null;
size--;
return rightNode;
}
if (node.right == null) {
Node leftNode = node.left;
node.left = null;
size--;
return leftNode;
}
Node successor = minimum(node.right);
successor.right = removeMin(node.right);
successor.left = node.left;
node.left = node.right = null;
return successor;
}
}
public V minimum() {
if (size == 0) {
throw new IllegalArgumentException("BST is empty!");
}
return minimum(root).value;
}
private Node minimum(Node node) {
if (node.left == null) {
return node;
}
return minimum(node.left);
}
public V removeMin() {
V ret = minimum();
root = removeMin(root);
return ret;
}
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;
}
@Override
public int getSize() {
return size;
}
@Override
public boolean isEmpty() {
return size == 0;
}
}