7-1 集合基础和基于二分搜索树的集合实现
07-Set-and-Map
集合
01-Set-Basics-and-BSTSet
文件IO操作，简单分词工具类
FileOperation
import java.io.FileInputStream;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.Locale;
import java.io.File;
import java.io.BufferedInputStream;
import java.io.IOException;
// 文件相关操作
public class FileOperation {
// 读取文件名称为filename中的内容，并将其中包含的所有词语放进words中
public static boolean readFile(String filename, ArrayList<String> words){
if (filename == null || words == null){
System.out.println("filename is null or words is null");
return false;
}
// 文件读取
Scanner scanner;
try {
File file = new File(filename);
if(file.exists()){
FileInputStream fis = new FileInputStream(file);
scanner = new Scanner(new BufferedInputStream(fis), "UTF-8");
scanner.useLocale(Locale.ENGLISH);
}
else
return false;
}
catch(IOException ioe){
System.out.println("Cannot open " + filename);
return false;
}
// 简单分词
// 这个分词方式相对简陋, 没有考虑很多文本处理中的特殊问题
// 在这里只做demo展示用
if (scanner.hasNextLine()) {
String contents = scanner.useDelimiter("\\A").next();
int start = firstCharacterIndex(contents, 0);
for (int i = start + 1; i <= contents.length(); )
if (i == contents.length() || !Character.isLetter(contents.charAt(i))) {
String word = contents.substring(start, i).toLowerCase();
words.add(word);
start = firstCharacterIndex(contents, i);
i = start + 1;
} else
i++;
}
return true;
}
// 寻找字符串s中，从start的位置开始的第一个字母字符的位置
private static int firstCharacterIndex(String s, int start){
for( int i = start ; i < s.length() ; i ++ )
if( Character.isLetter(s.charAt(i)) )
return i;
return s.length();
}
}
二分搜索树底层实现
BST.java
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
public class BST<E extends Comparable<E>> {
private class Node{
public E e;
public Node left, right;
public Node(E e){
this.e = e;
left = null;
right = null;
}
}
private Node root;
private int size;
public BST(){
root = null;
size = 0;
}
public int size(){
return size;
}
public boolean isEmpty(){
return size == 0;
}
// 向二分搜索树中添加新的元素e
public void add(E e){
root = add(root, e);
}
// 向以node为根的二分搜索树中插入元素e，递归算法
// 返回插入新节点后二分搜索树的根
private Node add(Node node, E e){
if(node == null){
size ++;
return new Node(e);
}
if(e.compareTo(node.e) < 0)
node.left = add(node.left, e);
else if(e.compareTo(node.e) > 0)
node.right = add(node.right, e);
return node;
}
// 看二分搜索树中是否包含元素e
public boolean contains(E e){
return contains(root, e);
}
// 看以node为根的二分搜索树中是否包含元素e, 递归算法
private boolean contains(Node node, E e){
if(node == null)
return false;
if(e.compareTo(node.e) == 0)
return true;
else if(e.compareTo(node.e) < 0)
return contains(node.left, e);
else // e.compareTo(node.e) > 0
return contains(node.right, e);
}
// 二分搜索树的前序遍历
public void preOrder(){
preOrder(root);
}
// 前序遍历以node为根的二分搜索树, 递归算法
private void preOrder(Node node){
if(node == null)
return;
System.out.println(node.e);
preOrder(node.left);
preOrder(node.right);
}
// 二分搜索树的非递归前序遍历
public void preOrderNR(){
Stack<Node> stack = new Stack<>();
stack.push(root);
while(!stack.isEmpty()){
Node cur = stack.pop();
System.out.println(cur.e);
if(cur.right != null)
stack.push(cur.right);
if(cur.left != null)
stack.push(cur.left);
}
}
// 二分搜索树的中序遍历
public void inOrder(){
inOrder(root);
}
// 中序遍历以node为根的二分搜索树, 递归算法
private void inOrder(Node node){
if(node == null)
return;
inOrder(node.left);
System.out.println(node.e);
inOrder(node.right);
}
// 二分搜索树的后序遍历
public void postOrder(){
postOrder(root);
}
// 后序遍历以node为根的二分搜索树, 递归算法
private void postOrder(Node node){
if(node == null)
return;
postOrder(node.left);
postOrder(node.right);
System.out.println(node.e);
}
// 二分搜索树的层序遍历
public void levelOrder(){
Queue<Node> q = new LinkedList<>();
q.add(root);
while(!q.isEmpty()){
Node cur = q.remove();
System.out.println(cur.e);
if(cur.left != null)
q.add(cur.left);
if(cur.right != null)
q.add(cur.right);
}
}
// 寻找二分搜索树的最小元素
public E minimum(){
if(size == 0)
throw new IllegalArgumentException("BST is empty!");
return minimum(root).e;
}
// 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);
}
// 寻找二分搜索树的最大元素
public E maximum(){
if(size == 0)
throw new IllegalArgumentException("BST is empty");
return maximum(root).e;
}
// 返回以node为根的二分搜索树的最大值所在的节点
private Node maximum(Node node){
if(node.right == null)
return node;
return maximum(node.right);
}
// 从二分搜索树中删除最小值所在节点, 返回最小值
public E removeMin(){
E ret = minimum();
root = removeMin(root);
return ret;
}
// 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最大值所在节点
public E removeMax(){
E ret = maximum();
root = removeMax(root);
return ret;
}
// 删除掉以node为根的二分搜索树中的最大节点
// 返回删除节点后新的二分搜索树的根
private Node removeMax(Node node){
if(node.right == null){
Node leftNode = node.left;
node.left = null;
size --;
return leftNode;
}
node.right = removeMax(node.right);
return node;
}
// 从二分搜索树中删除元素为e的节点
public void remove(E e){
root = remove(root, e);
}
// 删除掉以node为根的二分搜索树中值为e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, E e){
if( node == null )
return null;
if( e.compareTo(node.e) < 0 ){
node.left = remove(node.left , e);
return node;
}
else if(e.compareTo(node.e) > 0 ){
node.right = remove(node.right, e);
return node;
}
else{ // e.compareTo(node.e) == 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;
}
}
@Override
public String toString(){
StringBuilder res = new StringBuilder();
generateBSTString(root, 0, res);
return res.toString();
}
// 生成以node为根节点，深度为depth的描述二叉树的字符串
private void generateBSTString(Node node, int depth, StringBuilder res){
if(node == null){
res.append(generateDepthString(depth) + "null\n");
return;
}
res.append(generateDepthString(depth) + node.e +"\n");
generateBSTString(node.left, depth + 1, res);
generateBSTString(node.right, depth + 1, res);
}
private String generateDepthString(int depth){
StringBuilder res = new StringBuilder();
for(int i = 0 ; i < depth ; i ++)
res.append("--");
return res.toString();
}
}
Set集合接口
public interface Set<E> {
void add(E e);
boolean contains(E e);
void remove(E e);
int getSize();
boolean isEmpty();
}
BSTSet集合实现类
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);
}
}
测试主程序入口
Main.java
import java.util.ArrayList;
public class Main {
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());
BSTSet<String> set1 = new BSTSet<>();
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());
BSTSet<String> set2 = new BSTSet<>();
for(String word: words2)
set2.add(word);
System.out.println("Total different words: " + set2.getSize());
}
}
}
a-tale-of-two-cities.txt
pride-and-prejudice.txt
7-2 基于链表的集合实现
02-LinkedListSet
Set集合接口
public interface Set<E> {
void add(E e);
boolean contains(E e);
void remove(E e);
int getSize();
boolean isEmpty();
}
动态数据结构的底层实现
LinkedList
public class LinkedList<E> {
private class Node{
public E e;
public Node next;
public Node(E e, Node next){
this.e = e;
this.next = next;
}
public Node(E e){
this(e, null);
}
public Node(){
this(null, null);
}
@Override
public String toString(){
return e.toString();
}
}
private Node dummyHead;
private int size;
public LinkedList(){
dummyHead = new Node();
size = 0;
}
// 获取链表中的元素个数
public int getSize(){
return size;
}
// 返回链表是否为空
public boolean isEmpty(){
return size == 0;
}
// 在链表的index(0-based)位置添加新的元素e
// 在链表中不是一个常用的操作，练习用：）
public void add(int index, E e){
if(index < 0 || index > size)
throw new IllegalArgumentException("Add failed. Illegal index.");
Node prev = dummyHead;
for(int i = 0 ; i < index ; i ++)
prev = prev.next;
prev.next = new Node(e, prev.next);
size ++;
}
// 在链表头添加新的元素e
public void addFirst(E e){
add(0, e);
}
// 在链表末尾添加新的元素e
public void addLast(E e){
add(size, e);
}
// 获得链表的第index(0-based)个位置的元素
// 在链表中不是一个常用的操作，练习用：）
public E get(int index){
if(index < 0 || index >= size)
throw new IllegalArgumentException("Get failed. Illegal index.");
Node cur = dummyHead.next;
for(int i = 0 ; i < index ; i ++)
cur = cur.next;
return cur.e;
}
// 获得链表的第一个元素
public E getFirst(){
return get(0);
}
// 获得链表的最后一个元素
public E getLast(){
return get(size - 1);
}
// 修改链表的第index(0-based)个位置的元素为e
// 在链表中不是一个常用的操作，练习用：）
public void set(int index, E e){
if(index < 0 || index >= size)
throw new IllegalArgumentException("Set failed. Illegal index.");
Node cur = dummyHead.next;
for(int i = 0 ; i < index ; i ++)
cur = cur.next;
cur.e = e;
}
// 查找链表中是否有元素e
public boolean contains(E e){
Node cur = dummyHead.next;
while(cur != null){
if(cur.e.equals(e))
return true;
cur = cur.next;
}
return false;
}
// 从链表中删除index(0-based)位置的元素, 返回删除的元素
// 在链表中不是一个常用的操作，练习用：）
public E remove(int index){
if(index < 0 || index >= size)
throw new IllegalArgumentException("Remove failed. Index is illegal.");
Node prev = dummyHead;
for(int i = 0 ; i < index ; i ++)
prev = prev.next;
Node retNode = prev.next;
prev.next = retNode.next;
retNode.next = null;
size --;
return retNode.e;
}
// 从链表中删除第一个元素, 返回删除的元素
public E removeFirst(){
return remove(0);
}
// 从链表中删除最后一个元素, 返回删除的元素
public E removeLast(){
return remove(size - 1);
}
// 从链表中删除元素e
public void removeElement(E e){
Node prev = dummyHead;
while(prev.next != null){
if(prev.next.e.equals(e))
break;
prev = prev.next;
}
if(prev.next != null){
Node delNode = prev.next;
prev.next = delNode.next;
delNode.next = null;
size --;
}
}
@Override
public String toString(){
StringBuilder res = new StringBuilder();
Node cur = dummyHead.next;
while(cur != null){
res.append(cur + "->");
cur = cur.next;
}
res.append("NULL");
return res.toString();
}
}
集合接口的实现类
LinkedListSet
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());
}
}
}
7-3 集合类的复杂度分析
03-Time-Complexity-of-Set
Main
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");
}
}
7-4 Leetcode中的集合问题和更多集合相关问题
04-TreeSet-and-Set-Problems-in-Leetcode
// Leetcode 804. Unique Morse Code Words
// https://leetcode.com/problems/unique-morse-code-words/description/
LinkedListSetSolution
// Leetcode 804. Unique Morse Code Words
// https://leetcode.com/problems/unique-morse-code-words/description/
public class LinkedListSetSolution {
private class LinkedList<E> {
private class Node{
public E e;
public Node next;
public Node(E e, Node next){
this.e = e;
this.next = next;
}
public Node(E e){
this(e, null);
}
public Node(){
this(null, null);
}
@Override
public String toString(){
return e.toString();
}
}
private Node dummyHead;
private int size;
public LinkedList(){
dummyHead = new Node();
size = 0;
}
// 获取链表中的元素个数
public int getSize(){
return size;
}
// 返回链表是否为空
public boolean isEmpty(){
return size == 0;
}
// 在链表的index(0-based)位置添加新的元素e
// 在链表中不是一个常用的操作，练习用：）
public void add(int index, E e){
if(index < 0 || index > size)
throw new IllegalArgumentException("Add failed. Illegal index.");
Node prev = dummyHead;
for(int i = 0 ; i < index ; i ++)
prev = prev.next;
prev.next = new Node(e, prev.next);
size ++;
}
// 在链表头添加新的元素e
public void addFirst(E e){
add(0, e);
}
// 在链表末尾添加新的元素e
public void addLast(E e){
add(size, e);
}
// 获得链表的第index(0-based)个位置的元素
// 在链表中不是一个常用的操作，练习用：）
public E get(int index){
if(index < 0 || index >= size)
throw new IllegalArgumentException("Get failed. Illegal index.");
Node cur = dummyHead.next;
for(int i = 0 ; i < index ; i ++)
cur = cur.next;
return cur.e;
}
// 获得链表的第一个元素
public E getFirst(){
return get(0);
}
// 获得链表的最后一个元素
public E getLast(){
return get(size - 1);
}
// 修改链表的第index(0-based)个位置的元素为e
// 在链表中不是一个常用的操作，练习用：）
public void set(int index, E e){
if(index < 0 || index >= size)
throw new IllegalArgumentException("Set failed. Illegal index.");
Node cur = dummyHead.next;
for(int i = 0 ; i < index ; i ++)
cur = cur.next;
cur.e = e;
}
// 查找链表中是否有元素e
public boolean contains(E e){
Node cur = dummyHead.next;
while(cur != null){
if(cur.e.equals(e))
return true;
cur = cur.next;
}
return false;
}
// 从链表中删除index(0-based)位置的元素, 返回删除的元素
// 在链表中不是一个常用的操作，练习用：）
public E remove(int index){
if(index < 0 || index >= size)
throw new IllegalArgumentException("Remove failed. Index is illegal.");
Node prev = dummyHead;
for(int i = 0 ; i < index ; i ++)
prev = prev.next;
Node retNode = prev.next;
prev.next = retNode.next;
retNode.next = null;
size --;
return retNode.e;
}
// 从链表中删除第一个元素, 返回删除的元素
public E removeFirst(){
return remove(0);
}
// 从链表中删除最后一个元素, 返回删除的元素
public E removeLast(){
return remove(size - 1);
}
// 从链表中删除元素e
public void removeElement(E e){
Node prev = dummyHead;
while(prev.next != null){
if(prev.next.e.equals(e))
break;
prev = prev.next;
}
if(prev.next != null){
Node delNode = prev.next;
prev.next = delNode.next;
delNode.next = null;
size --;
}
}
@Override
public String toString(){
StringBuilder res = new StringBuilder();
Node cur = dummyHead.next;
while(cur != null){
res.append(cur + "->");
cur = cur.next;
}
res.append("NULL");
return res.toString();
}
}
private interface Set<E> {
void add(E e);
boolean contains(E e);
void remove(E e);
int getSize();
boolean isEmpty();
}
private 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 int uniqueMorseRepresentations(String[] words) {
String[] codes = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
LinkedListSet<String> set = new LinkedListSet<>();
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.getSize();
}
}
BSTSetSolution
// Leetcode 804. Unique Morse Code Words
// https://leetcode.com/problems/unique-morse-code-words/description/
import java.util.Stack;
import java.util.Queue;
import java.util.LinkedList;
public class BSTSetSolution {
private class BST<E extends Comparable<E>> {
private class Node{
public E e;
public Node left, right;
public Node(E e){
this.e = e;
left = null;
right = null;
}
}
private Node root;
private int size;
public BST(){
root = null;
size = 0;
}
public int size(){
return size;
}
public boolean isEmpty(){
return size == 0;
}
// 向二分搜索树中添加新的元素e
public void add(E e){
root = add(root, e);
}
// 向以node为根的二分搜索树中插入元素e，递归算法
// 返回插入新节点后二分搜索树的根
private Node add(Node node, E e){
if(node == null){
size ++;
return new Node(e);
}
if(e.compareTo(node.e) < 0)
node.left = add(node.left, e);
else if(e.compareTo(node.e) > 0)
node.right = add(node.right, e);
return node;
}
// 看二分搜索树中是否包含元素e
public boolean contains(E e){
return contains(root, e);
}
// 看以node为根的二分搜索树中是否包含元素e, 递归算法
private boolean contains(Node node, E e){
if(node == null)
return false;
if(e.compareTo(node.e) == 0)
return true;
else if(e.compareTo(node.e) < 0)
return contains(node.left, e);
else // e.compareTo(node.e) > 0
return contains(node.right, e);
}
// 二分搜索树的前序遍历
public void preOrder(){
preOrder(root);
}
// 前序遍历以node为根的二分搜索树, 递归算法
private void preOrder(Node node){
if(node == null)
return;
System.out.println(node.e);
preOrder(node.left);
preOrder(node.right);
}
// 二分搜索树的非递归前序遍历
public void preOrderNR(){
Stack<Node> stack = new Stack<>();
stack.push(root);
while(!stack.isEmpty()){
Node cur = stack.pop();
System.out.println(cur.e);
if(cur.right != null)
stack.push(cur.right);
if(cur.left != null)
stack.push(cur.left);
}
}
// 二分搜索树的中序遍历
public void inOrder(){
inOrder(root);
}
// 中序遍历以node为根的二分搜索树, 递归算法
private void inOrder(Node node){
if(node == null)
return;
inOrder(node.left);
System.out.println(node.e);
inOrder(node.right);
}
// 二分搜索树的后序遍历
public void postOrder(){
postOrder(root);
}
// 后序遍历以node为根的二分搜索树, 递归算法
private void postOrder(Node node){
if(node == null)
return;
postOrder(node.left);
postOrder(node.right);
System.out.println(node.e);
}
// 二分搜索树的层序遍历
public void levelOrder(){
Queue<Node> q = new LinkedList<>();
q.add(root);
while(!q.isEmpty()){
Node cur = q.remove();
System.out.println(cur.e);
if(cur.left != null)
q.add(cur.left);
if(cur.right != null)
q.add(cur.right);
}
}
// 寻找二分搜索树的最小元素
public E minimum(){
if(size == 0)
throw new IllegalArgumentException("BST is empty!");
return minimum(root).e;
}
// 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);
}
// 寻找二分搜索树的最大元素
public E maximum(){
if(size == 0)
throw new IllegalArgumentException("BST is empty");
return maximum(root).e;
}
// 返回以node为根的二分搜索树的最大值所在的节点
private Node maximum(Node node){
if(node.right == null)
return node;
return maximum(node.right);
}
// 从二分搜索树中删除最小值所在节点, 返回最小值
public E removeMin(){
E ret = minimum();
root = removeMin(root);
return ret;
}
// 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最大值所在节点
public E removeMax(){
E ret = maximum();
root = removeMax(root);
return ret;
}
// 删除掉以node为根的二分搜索树中的最大节点
// 返回删除节点后新的二分搜索树的根
private Node removeMax(Node node){
if(node.right == null){
Node leftNode = node.left;
node.left = null;
size --;
return leftNode;
}
node.right = removeMax(node.right);
return node;
}
// 从二分搜索树中删除元素为e的节点
public void remove(E e){
root = remove(root, e);
}
// 删除掉以node为根的二分搜索树中值为e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, E e){
if( node == null )
return null;
if( e.compareTo(node.e) < 0 ){
node.left = remove(node.left , e);
return node;
}
else if(e.compareTo(node.e) > 0 ){
node.right = remove(node.right, e);
return node;
}
else{ // e.compareTo(node.e) == 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;
}
}
@Override
public String toString(){
StringBuilder res = new StringBuilder();
generateBSTString(root, 0, res);
return res.toString();
}
// 生成以node为根节点，深度为depth的描述二叉树的字符串
private void generateBSTString(Node node, int depth, StringBuilder res){
if(node == null){
res.append(generateDepthString(depth) + "null\n");
return;
}
res.append(generateDepthString(depth) + node.e +"\n");
generateBSTString(node.left, depth + 1, res);
generateBSTString(node.right, depth + 1, res);
}
private String generateDepthString(int depth){
StringBuilder res = new StringBuilder();
for(int i = 0 ; i < depth ; i ++)
res.append("--");
return res.toString();
}
}
private interface Set<E> {
void add(E e);
boolean contains(E e);
void remove(E e);
int getSize();
boolean isEmpty();
}
private 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);
}
}
public int uniqueMorseRepresentations(String[] words) {
String[] codes = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
BSTSet<String> set = new BSTSet<>();
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.getSize();
}
}
底层是平衡二叉树（红黑树）
Solution.java
// 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();
}
}
原文地址：https://www.cnblogs.com/MarlonKang/p/12362335.html
时间： 2024-09-29 09:27:23