1.. 集合的应用
- 集合可以用来去重
- 集合可以用于进行客户的统计
- 集合可以用于文本词汇量的统计
2.. 集合的实现
- 定义集合的接口
Set<E>
·void add(E) // 不能添加重复元素
·void remove(E)
·boolean contains(E)
·int getSize()
·boolean isEmpty()集合接口的业务逻辑如下：
public interface Set<E> { void add(E e); void remove(E e); boolean contains(E e); int getSize(); boolean isEmpty();
}- 用二分搜索树作为集合的底层实现
public class BSTSet<E extends Comparable<E>> implements Set<E> { private BST<E> bst; // 构造函数
public BSTSet() {
bst = new BST<>();
} // 实现getSize方法
@Override
public int getSize() {
return bst.size();
} // 实现isEmpty方法
@Override
public boolean isEmpty() {
return bst.isEmpty();
} // 实现contains方法
@Override
public boolean contains(E e) {
return bst.contains(e);
} // 实现add方法
public void add(E e) {
bst.add(e);
} // 实现remove方法
public void remove(E e) {
bst.remove(e);
}
}- 用链表作为集合的底层实现
public class LinkedListSet<E> implements Set<E> { private LinkedList<E> list; // 构造函数
public LinkedListSet() {
list = new LinkedList<>();
} // 实现getSize方法
@Override
public int getSize() {
return list.getSize();
} // 实现isEmpty方法
@Override
public boolean isEmpty() {
return list.isEmpty();
} // 实现contains方法
@Override
public boolean contains(E e) {
return list.contains(e);
} // 实现add方法
@Override
public void add(E e) {
if (!list.contains(e)) {
list.addFirst(e);
}
} // 实现remove方法
@Override
public void remove(E e) {
list.removeElement(e);
}- 用二分搜索树实现的集合与用链表实现的集合的性能比较
import java.util.ArrayList; public class Main { public 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;
} public static void main(String[] args) {
String filename = "pride-and-prejudice.txt"; BSTSet<String> bstSet = new BSTSet<>();
double time1 = testSet(bstSet, filename);
System.out.println("BSTSet, time: " + time1 + " s"); System.out.println(); LinkedListSet<String> linkedListSet = new LinkedListSet<>();
double time2 = testSet(linkedListSet, filename);
System.out.println("LinkedListSet, time: " + time2 + " s");
}
}- 输出结果：
pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
BSTSet, time: 0.109504342 s pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
LinkedListSet, time: 2.208894105 s
- 通过比较结果，我们发现，用二分搜索树实现的集合的比用链表实现的集合更加高效
3.. 集合的时间复杂度分析
- 上图中"h"是二分搜索树的高度
- 当二分搜索树"满"的时候，性能是最佳的，时间复杂度为O(logn)；当二分搜索树退化为链表的时候，性能是最差的，时间复杂度为O(n)
4.. 映射（Map）
- 映射是存储（键，值）数据对的数据结构（Key, Value）
- 根据键（Key），寻找值（Value）
5.. 映射的实现
- 定义映射的接口
Map<K, V>
·void add(K, V)
·V remove(K)
·boolean contains(K)
·V get(K)
·void set(K, V)
·int getSize()
·boolean isEmpty()- 映射接口的业务逻辑如下
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 value); int getSize(); boolean isEmpty();
}- 用链表作为映射的底层实现
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; // 构造函数
public LinkedListMap() {
dummyHead = new Node();
size = 0;
} // 实现getSize方法
@Override
public int getSize() {
return size;
} // 实现isEmpty方法
@Override
public boolean isEmpty() {
return size == 0;
} private Node getNode(K key) {
Node cur = dummyHead;
while (cur != null) {
if (cur.key.equals(key)) {
return cur;
}
cur = cur.next;
}
return null;
} // 实现contains方法
@Override
public boolean contains(K key) {
return getNode(key) != null;
} // 实现get方法
@Override
public V get(K key) {
Node node = getNode(key); // return node == null ? null : node.value;
if (node != null) {
return node.value;
}
return null;
} // 实现add方法
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;
}
} // 实现set方法
public void set(K key, V newValue) {
Node node = getNode(key);
if (node == null) {
throw new IllegalArgumentException(key + " doesn't exist.");
} else {
node.value = newValue;
}
} // 实现remove方法
public V remove(K key) { Node node = getNode(key);
if (node == null) {
throw new IllegalArgumentException(key + " doesn't exist.");
} 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;
}
}- 用二分搜索树作为映射的底层实现
public class BSTMap<K extends Comparable<K>, V> implements Map<K, V> { private class Node {
private K key;
private V value;
private Node left;
private Node right; // 构造函数
public Node(K key, V value) {
this.key = key;
this.value = value;
this.left = null;
this.right = null;
} // public Node(K key) {
// this(key, null);
// }
} private Node root;
private int size; // 构造函数
public BSTMap() {
root = null;
size = 0;
} // 实现getSize方法
@Override
public int getSize() {
return size;
} // 实现isEmpty方法
public boolean isEmpty() {
return size == 0;
} // 实现add方法
@Override
public void add(K key, V value) {
root = add(root, key, value);
} // 向以node为根节点的二分搜索树中插入元素(key, value)，递归算法
// 返回插入新元素后的二分搜索树的根
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 {
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.compareTo(node.key) < 0) {
return getNode(node.left, key);
} else if (key.compareTo(node.key) > 0) {
return getNode(node.right, key);
} else {
return node;
}
} @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元素所在节点
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元素所在节点
// 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;
} // 实现remove方法
// 删除二分搜索树中键为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;
} // 删除以node为根节点的二分搜索树中键为key的节点，递归算法
// 返回删除节点后新的二分搜索树的根
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 {
// 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
} else 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
// 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
} else {
Node successor = minimum(node.right);
successor.right = removeMin(node.right); //这里进行了size--操作
successor.left = node.left;
node.left = null;
node.right = null;
return successor;
}
}
}
}- 用二分搜索树实现的映射与用链表实现的映射的性能比较
import java.util.ArrayList; public class Main { public static double testMap(Map<String, Integer> map, 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) {
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"));
} long endTime = System.nanoTime(); return (endTime - startTime) / 1000000000.0;
} public static void main(String[] args) { String filename = "pride-and-prejudice.txt"; LinkedListMap<String, Integer> linkedListMap = new LinkedListMap<>();
double time1 = testMap(linkedListMap, filename);
System.out.println("Linked List Map, time: " + time1 + " s"); System.out.println();
System.out.println(); BSTMap<String, Integer> bstMap = new BSTMap<>();
double time2 = testMap(bstMap, filename);
System.out.println("BST Map, time: " + time2 + " s"); }
}- 输出结果
pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
Frequency of PRIDE: 53
Frequency of PREJUDICE: 11
Linked List Map, time: 9.692566895 s pride-and-prejudice.txt
Total words: 125901
Total different words: 6530
Frequency of PRIDE: 53
Frequency of PREJUDICE: 11
BST Map, time: 0.085364242 s- 通过比较结果，我们发现，用二分搜索树实现的映射的比用链表实现的映射更加高效
6.. 映射的时间复杂度
- 上图中"h"是二分搜索树的高度
- 当二分搜索树"满"的时候，性能是最佳的，时间复杂度为O(logn)；当二分搜索树退化为链表的时候，性能是最差的，时间复杂度为O(n)