集合
以二分搜索树为底层的集合实现
基于上节的二分搜索树
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);
}
}
以链表为底层的集合实现
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);
}
public static void main(String[] args) {
System.out.println("Pride and Prejudice");
ArrayList<String> words1 = new ArrayList<>();
if(FileOperation.readFile("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());
}
System.out.println();
System.out.println("A Tale of Two Cities");
ArrayList<String> words2 = new ArrayList<>();
if(FileOperation.readFile("a-tale-of-two-cities.txt", words2)){
System.out.println("Total words: " + words2.size());
LinkedListSet<String> set2 = new LinkedListSet<>();
for(String word: words2)
set2.add(word);
System.out.println("Total different words: " + set2.getSize());
}
}
}
时间复杂度测试
当二叉树只有一条的时候，时间复杂度和链表的相同。
import java.util.ArrayList;
public class Main {
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;
}
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("BST Set: " + time1 + " s");
System.out.println();
LinkedListSet<String> linkedListSet = new LinkedListSet<>();
double time2 = testSet(linkedListSet, filename);
System.out.println("Linked List Set: " + time2 + " s");
}
}LeetCode 804 唯一摩尔斯密码词
// Leetcode 804. Unique Morse Code Words
// https://leetcode.com/problems/unique-morse-code-words/description/
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();
}
}
映射（字典）Map
以链表为底层的映射实现
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;
// 循环
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;
}
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());
LinkedListMap<String, Integer> map = new LinkedListMap<>();
for (String word : words) {
if (map.contains(word))
// 频率加1
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();
}
}
以二分搜索树为底层的映射实现
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();
}
}
Test
import java.util.ArrayList;
public class Main {
public static void main(String[] args) {
ArrayList<String> words = new ArrayList<>();
if(FileOperation.readFile("pride-and-prejudice.txt", words) &&
FileOperation.readFile("a-tale-of-two-cities.txt", words)){
// Test BST
long startTime = System.nanoTime();
BSTSet<String> set = new BSTSet<>();
for(String word: words)
set.add(word);
for(String word: words)
set.contains(word);
long endTime = System.nanoTime();
double time = (endTime - startTime) / 1000000000.0;
System.out.println("Total different words: " + set.getSize());
System.out.println("BSTSet: " + time + " s");
// ---
// Test TreeMap Trie
startTime = System.nanoTime();
Trie trie = new Trie();
for(String word: words)
trie.add(word);
for(String word: words)
trie.contains(word);
endTime = System.nanoTime();
time = (endTime - startTime) / 1000000000.0;
System.out.println("Total different words: " + trie.getSize());
System.out.println("TreeMap Trie: " + time + " s");
// ---
// Test HashMap Trie
startTime = System.nanoTime();
Trie2 trie2 = new Trie2();
for(String word: words)
trie2.add(word);
for(String word: words)
trie2.contains(word);
endTime = System.nanoTime();
time = (endTime - startTime) / 1000000000.0;
System.out.println("Total different words: " + trie.getSize());
System.out.println("HashMap Trie: " + time + " s");
// ---
// Test Array(Map) Trie
startTime = System.nanoTime();
Trie3 trie3 = new Trie3();
for(String word: words)
trie3.add(word);
for(String word: words)
trie3.contains(word);
endTime = System.nanoTime();
time = (endTime - startTime) / 1000000000.0;
System.out.println("Total different words: " + trie.getSize());
System.out.println("Array(Map) Trie: " + time + " s");
}
}
}
Leetcode 349 两个数组的交集 使用集合
Leetcode 350 两个数组的交集 使用映射
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