数组结构七：集合和映射（Set And Map）
}
1.2、BSTSet.java
public class BSTSet<E extends Comparable<E>> implements Set<E> {
private BST<E> bst;
public BSTSet() {
bst = new BST<>();
}
@Override
public void add(E e) {
bst.add(e);
}
@Override
public void remove(E e) {
bst.remove(e);
}
@Override
public boolean contains(E e) {
return bst.contains(e);
}
@Override
public int getSize() {
return bst.size();
}
@Override
public boolean isEmpty() {
return false;
}
public static void main(String[] args) {
System.out.println("Pride and Prejudice");
ArrayList<String> words1 = new ArrayList<>();
FileOperation.readFile(
"D:\idea_ws\java\JavaTest1\src\com\ph\pride-and-prejudice.txt", words1);
// FileOperation.readFile(
// "..\..\..\pride-and-prejudice.txt", words1);
System.out.println("Total words: " + words1.size());
BSTSet<String> set1=new BSTSet<>();
for (String word:words1)
set1.add(word);
System.out.println("Total different words: " + set1.getSize());
}
}
1.3、LinkedListSet.java
public class LinkedListSet<E> implements Set<E> {
private LearnLinkedList<E> list;
public LinkedListSet() {
list = new LearnLinkedList<>();
}
/**
LearnLinkedList 可以有重复的元素
Set 不可以有重复的元素
所以这里要稍作一些逻辑的处理
*/
@Override
public void add(E e) {
if (!list.contains(e))
list.addFirst(e);
}
@Override
public void remove(E e) {
list.removeElement(e);
}
@Override
public boolean contains(E e) {
return list.contains(e);
}
@Override
public int getSize() {
return list.getSize();
}
@Override
public boolean isEmpty() {
return list.isEmpty();
}
public static void main(String[] args) {
System.out.println("Pride and Prejudice");
ArrayList<String> words1 = new ArrayList<>();
FileOperation.readFile(
"D:\idea_ws\java\JavaTest1\src\com\ph\pride-and-prejudice.txt", words1);
System.out.println("Total words: " + words1.size());
LinkedListSet<String> set1 = new LinkedListSet<>();
for (String word : words1)
set1.add(word);
System.out.println("Total different words: " + set1.getSize());
}
}
1.4、SetMainTest.java
/**
测试 BSTSet 和 LinkedListSet 的性能
*/
public class SetMainTest {
public static void main(String[] args) {
String fileName = "D:\idea_ws\java\JavaTest1\src\com\ph\pride-and-prejudice.txt";
double time1 = testSet(new BSTSet<>(), fileName);
System.out.println("BSTSet: " + time1 + "s");
System.out.println();
double time2 = testSet(new LinkedListSet<>(), fileName);
System.out.println("LinkedListSet: " + time2 + "s");
}
private static double testSet(Set<String> set, String fileName) {
long startTime = System.nanoTime();
System.out.println(fileName);
ArrayList<String> words = new ArrayList<>();
if (FileOperation.readFile(fileName, words)) {
System.out.println("Total words: " + words.size());
for (String word : words)
set.add(word);
System.out.println("Total different words: " + set.getSize());
}
long endTime = System.nanoTime();
return (endTime - startTime) / 1000000000.0;
}
}
打印结果：
可以看到 BST 二分搜索树的性能要远高于 LinkedListSet。
下面我们进行时间复杂度分析。
2、集合的时间复杂度分析
============
其中的 h 是指二分搜索树的高度，那么 h 和 n 的关系是怎么样的呢？
如下：
分析 1：
分析 2：
分析 3：
分析 4：
假设数据量为 100 万，如果一个 O（logn）算法需要一天的话，O（n）算法需要 137 年。
分析 5：
3、Map?映射
========
称映射不是那么好理解，称为字典就更好理解一点。在 Python 中，map 这种数据结构的名字就是 dict（dictionary 的缩写?字典）、
总结：
存储（键，值）数据对的数据结构（Key，Value）
根据 Key，寻找 Value
非常容易使用链表或者二分搜索树实现
3.1、Map.java
public interface Map<K, V> {
int getSize();
boolean isEmpty();
void add(K key, V value);
V remove(K key);
boolean contains(K key);
V get(K key);
void set(K key, V newValue);
}
3.2、LinkedListMap.java
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;
}
@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;
}
/**
不能有两个相同的 key
*/
@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 + "不存在！");
node.value = newValue;
}
// 最复杂
@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;
}
public static void main(String[] args) {
System.out.println("Pride and Prejudice");
ArrayList<String> words1 = new ArrayList<>();
if (FileOperation.readFile(Constant.FILE_NAME, words1)) {
System.out.println("Total words: " + words1.size());
// 用来做单词频率统计，string 用于放单词，integer 用于放频率
LinkedListMap<String, Integer> map = new LinkedListMap<>();
for (String word : words1) {
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"));
}
}
}
3.3、BSTMap.java
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;
@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) {
siz
e++;
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;
}
@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;
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