集合(set)
特点
每个元素只能存在一个，可基于二分搜索树或者链表实现
基于二分搜索树的实现
二分搜索树代码
package com.study.set;
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();
}
}
接口代码
package com.study.set;
public interface Set<E> {
void add(E e);
boolean contains(E e);
void remove(E e);
int getSize();
boolean isEmpty();
}
treeSet代码
package com.study.set;
/**
* Created by Zsy on 2020/8/14.
*/
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 boolean contains(E e) {
return bst.contains(e);
}
@Override
public void remove(E e) {
bst.remove(e);
}
@Override
public int getSize() {
return bst.size();
}
@Override
public boolean isEmpty() {
return bst.isEmpty();
}
}
基于链表实现集合
链表
package com.study.set;
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();
}
}
linklistSet
package com.study.set;
/**
* Created by Zsy on 2020/8/14.
*/
public class LinkedListSet<E> implements Set<E> {
private LinkedList linkedList;
public LinkedListSet() {
linkedList = new LinkedList<>();
}
@Override
public void add(E e) {
if (!linkedList.contains(e))
linkedList.addFirst(e);
}
@Override
public boolean contains(E e) {
return linkedList.contains(e);
}
@Override
public void remove(E e) {
linkedList.removeElement(e);
}
@Override
public int getSize() {
return linkedList.getSize();
}
@Override
public boolean isEmpty() {
return linkedList.isEmpty();
}
}
两种方式实现集合发的复杂度分析
注意：二分搜索树实现的set有可能会退化成链表，此时的时间复杂度为O(n)
linklistSet bstSet
add O(n) O(logn)
remove O(n) O(logn)
contain O(n) O(logn)
补充：若要实现有序集合，建议用哈希表再封装为set
leetcode中关于集合的题目
804. 唯一摩尔斯密码词
解决思路
1. 声明单词对应摩斯码
2. 遍历单词，并将每个单词对应字母寻找对应摩斯码进行拼接
3. 用set集合进行添加，若为重复set就不会添加，问题得以解决
实现代码
public class uniqueMorseRepresentations_804 {
public int uniqueMorseRepresentations(String[] words) {
String[] codes = {
".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
TreeSet<String> treeSet=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']);
}
treeSet.add(res.toString()) ;
}
return treeSet.size();
}
}
映射
概述
数学中的函数就是映射，一一对应，俗称字典
实现
基于二分搜索树
基于链表
package com.study.map;
/**
* Created by Zsy on 2020/8/14.
*/
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;
}
public Node getNode(K key) {
Node cur = dummyHead.next;
while (cur != null) {
if (key.equals(cur.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 V remove(K key) {
Node prev = dummyHead;
while (prev.next != null) {
if (key.equals(prev.next.key)) {
break;
}
prev = prev.next;
}
if (prev.next != null) {
Node delNode = prev.next;
prev.next = delNode.next;
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 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 int getSize() {
return size;
}
@Override
public boolean isEmpty() {
return size == 0;
}
}
基于二分搜索树
package com.study.map;
/**
* Created by Zsy on 2020/8/14.
*/
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) {
}
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.equals(node.key))
return node;
else if (key.compareTo(node.key) < 0)
return getNode(node.left, key);
else
return getNode(node.right, key);
}
@Override
public V remove(K key) {
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;
}
}
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;
}
private Node minimum(Node node) {
if (node.left == null)
return node;
return minimum(node.left);
}
@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;
}
}
复杂度分析
注意：二分搜索树实现的set有可能会退化成链表，此时的时间复杂度为O(n)
linklistSet bstSet
add O(n) O(logn)
remove O(n) O(logn)
contain O(n) O(logn)
映射和集合的关系
集合包含于映射，可以用映射实现集合
leetcode中映射和集合的使用场景分门别类讲解
349 两个数组交集1
链接
解题思路
这题要求求出两个数组交集，且交集中元素不重复，所以解题步骤为：
1. 使用set遍历存储数组1元素
2. 遍历数组2，若遍历的元素在数组1中也存在，就添加至arraylist中，同时注意数组1此时得移除与数组2相同的元素，避免list重复添加
实现代码
public int[] intersection(int[] nums1, int[] nums2) {
TreeSet<Integer> set = new TreeSet<>();
for (int num : nums1) {
set.add(num);
}
ArrayList<Integer> list = new ArrayList<>();
for (int num : nums2) {
if (set.contains(num)) {
list.add(num);
set.remove(num);
}
}
int[] res = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
res[i] = list.get(i);
}
return res;
}
350. 两个数组的交集 II
链接
解题思路
与上一题不同的是，本题交集允许重复，所以建议使用map来解决问题，具体步骤为:
1. 遍历nums1，若map中存在遍历元素，则该元素value+1，不存在则创建以遍历元素为key的映射
2. 遍历nums2，这里设遍历元素名为key，若map中存在key，就将key存入list中，并将map中的key对应的value-1，注意当key的value等于0时移除map中的这个key，并list添加过多的key，造成错误。
实现代码
public int[] intersect(int[] nums1, int[] nums2) {
TreeMap<Integer, Integer> map = new TreeMap<>();
for (int num : nums1) {
if (map.containsKey(num)) {
map.put(num, map.get(num) + 1);
} else {
map.put(num, 1);
}
}
ArrayList<Integer> list = new ArrayList<>();
for (int num : nums2) {
if (map.containsKey(num)) {
list.add(num);
map.put(num, map.get(num) - 1);
if (map.get(num)==0)
map.remove(num);
}
}
int[] res = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
res[i] = list.get(i);
}
return res;
}