集合(Set)
◼ 集合的特点
不存放重复的元素
常用于去重
✓存放新增 IP，统计新增 IP 量
✓ 存放词汇，统计词汇量
✓ ……
public interface Set<E> {
int size();
boolean isEmpty();
void clear();
boolean contains(E element);
void add(E element);
void remove(E element);
void traversal(Visitor<E> visitor);// 遍历
public static abstract class Visitor<E> {
boolean stop;
public abstract boolean visit(E element);
}
}
◼ 思考:集合的内部实现能否直接利用以前学过的数据结构?
动态数组
链表
二叉搜索树(AVL树、红黑树)
链表实现集合
package alangeit.set;
// 链表实现集合
import alangeit.list.LinkedList;
import alangeit.list.List;
public class ListSet<E> implements Set<E> {
private List<E> list = new LinkedList<>();
@Override
public int size() {
return list.size();
}
@Override
public boolean isEmpty() {
return list.isEmpty();
}
@Override
public void clear() {
list.clear();
}
@Override
public boolean contains(E element) {
return list.contains(element);
}
@Override
public void add(E element) {
int index = list.indexOf(element);
if (index != List.ELEMENT_NOT_FOUND) { // 存在就覆盖
list.set(index, element);
} else { // 不存在就添加
list.add(element);
}
}
@Override
public void remove(E element) {
int index = list.indexOf(element);
if (index != List.ELEMENT_NOT_FOUND) {
list.remove(index);
}
}
@Override
public void traversal(Visitor<E> visitor) {
if (visitor == null) return;
int size = list.size();
for (int i = 0; i < size; i++) {
if (visitor.visit(list.get(i))) return;
}
}
}
红黑树实现集合
package alangeit.set;
// 红黑树实现集合
import java.util.Comparator;
import alangeit.tree.BinaryTree;
import alangeit.tree.RBTree;
public class TreeSet<E> implements Set<E> {
private RBTree<E> tree;
public TreeSet() {
this(null);
}
public TreeSet(Comparator<E> comparator) {
tree = new RBTree<>(comparator);
}
@Override
public int size() {
return tree.size();
}
@Override
public boolean isEmpty() {
return tree.isEmpty();
}
@Override
public void clear() {
tree.clear();
}
@Override
public boolean contains(E element) {
return tree.contains(element);
}
@Override
public void add(E element) {
tree.add(element);
}
@Override
public void remove(E element) {
tree.remove(element);
}
@Override
public void traversal(Visitor<E> visitor) {
tree.inorder(new BinaryTree.Visitor<E>() {
@Override
public boolean visit(E element) {
return visitor.visit(element);
}
});
}
}
// 红黑树实现集合
static void test2() {
System.out.println("--------------------------------------- 红黑树实现集合");
Set<Integer> treeSet = new TreeSet<>();
treeSet.add(12);
treeSet.add(10);
treeSet.add(7);
treeSet.add(11);
treeSet.add(10);
treeSet.add(11);
treeSet.add(9);
treeSet.traversal(new Visitor<Integer>() {
@Override
public boolean visit(Integer element) {
System.out.println(element);
return false;
}
});
}
作业
◼ 两个数组的交集:https://leetcode-cn.com/problems/intersection-of-two-arrays/
映射(Map)
◼Map 在有些编程语言中也叫做字典(dictionary，比如 Python、Objective-C、Swift 等) key value
◼Map 的每一个 key 是唯一的
Map的接口设计
public interface Map<K, V> {
int size();
boolean isEmpty();
void clear();
V put(K key, V value);// 添加
V get(K key);
V remove(K key);
boolean containsKey(K key);
boolean containsValue(V value);
void traversal(Visitor<K, V> visitor);// 遍历
public static abstract class Visitor<K, V> {
boolean stop;
public abstract boolean visit(K key, V value);
}
}
◼类似 Set，Map 可以直接利用之前学习的链表、二叉搜索树(AVL树、红黑树)等数据结构来实现
Map.java
package alangeit.map;
public interface Map<K, V> {
int size();
boolean isEmpty();
void clear();
V put(K key, V value);// 添加
V get(K key);
V remove(K key);
boolean containsKey(K key);
boolean containsValue(V value);
void traversal(Visitor<K, V> visitor);// 遍历
public static abstract class Visitor<K, V> {
boolean stop;
public abstract boolean visit(K key, V value);
}
}
TreeMap.java
package alangeit.map;
// 红黑树实现映射
import java.util.Comparator;
import java.util.LinkedList;
import java.util.Queue;
@SuppressWarnings({"unchecked", "unused"})
public class TreeMap<K, V> implements Map<K, V> {
private static final boolean RED = false;
private static final boolean BLACK = true;
private int size;
private Node<K, V> root;
private Comparator<K> comparator;
public TreeMap() {
this(null);
}
public TreeMap(Comparator<K> comparator) {
this.comparator = comparator;
}
public int size() {
return size;
}
public boolean isEmpty() {
return size == 0;
}
public void clear() {
root = null;
size = 0;
}
@Override
public V put(K key, V value) {
keyNotNullCheck(key);
// 添加第一个节点
if (root == null) {
root = new Node<>(key, value, null);
size++;
// 新添加节点之后的处理
afterPut(root);
return null;
}
// 添加的不是第一个节点
// 找到父节点
Node<K, V> parent = root;
Node<K, V> node = root;
int cmp = 0;
do {
cmp = compare(key, node.key);
parent = node;
if (cmp > 0) {
node = node.right;
} else if (cmp < 0) {
node = node.left;
} else { // 相等
node.key = key;
V oldValue = node.value;
node.value = value;
return oldValue;
}
} while (node != null);
// 看看插入到父节点的哪个位置
Node<K, V> newNode = new Node<>(key, value, parent);
if (cmp > 0) {
parent.right = newNode;
} else {
parent.left = newNode;
}
size++;
// 新添加节点之后的处理
afterPut(newNode);
return null;
}
@Override
public V get(K key) {
Node<K, V> node = node(key);
return node != null ? node.value : null;
}
@Override
public V remove(K key) {
return remove(node(key));
}
@Override
public boolean containsKey(K key) {
return node(key) != null;
}
@Override
public boolean containsValue(V value) {
if (root == null) return false;
Queue<Node<K, V>> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
Node<K, V> node = queue.poll();
if (valEquals(value, node.value)) return true;
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
return false;
}
@Override
public void traversal(Visitor<K, V> visitor) {
if (visitor == null) return;
traversal(root, visitor);
}
private void traversal(Node<K, V> node, Visitor<K, V> visitor) {
if (node == null || visitor.stop) return;
traversal(node.left, visitor);
if (visitor.stop) return;
visitor.visit(node.key, node.value);
traversal(node.right, visitor);
}
private boolean valEquals(V v1, V v2) {
return v1 == null ? v2 == null : v1.equals(v2);
}
private V remove(Node<K, V> node) {
if (node == null) return null;
size--;
V oldValue = node.value;
if (node.hasTwoChildren()) { // 度为2的节点
// 找到后继节点
Node<K, V> s = successor(node);
// 用后继节点的值覆盖度为2的节点的值
node.key = s.key;
node.value = s.value;
// 删除后继节点
node = s;
}
// 删除node节点（node的度必然是1或者0）
Node<K, V> replacement = node.left != null ? node.left : node.right;
if (replacement != null) { // node是度为1的节点
// 更改parent
replacement.parent = node.parent;
// 更改parent的left、right的指向
if (node.parent == null) { // node是度为1的节点并且是根节点
root = replacement;
} else if (node == node.parent.left) {
node.parent.left = replacement;
} else { // node == node.parent.right
node.parent.right = replacement;
}
// 删除节点之后的处理
afterRemove(replacement);
} else if (node.parent == null) { // node是叶子节点并且是根节点
root = null;
} else { // node是叶子节点，但不是根节点
if (node == node.parent.left) {
node.parent.left = null;
} else { // node == node.parent.right
node.parent.right = null;
}
// 删除节点之后的处理
afterRemove(node);
}
return oldValue;
}
private void afterRemove(Node<K, V> node) {
// 如果删除的节点是红色
// 或者 用以取代删除节点的子节点是红色
if (isRed(node)) {
black(node);
return;
}
Node<K, V> parent = node.parent;
if (parent == null) return;
// 删除的是黑色叶子节点【下溢】
// 判断被删除的node是左还是右
boolean left = parent.left == null || node.isLeftChild();
Node<K, V> sibling = left ? parent.right : parent.left;
if (left) { // 被删除的节点在左边，兄弟节点在右边
if (isRed(sibling)) { // 兄弟节点是红色
black(sibling);
red(parent);
rotateLeft(parent);
// 更换兄弟
sibling = parent.right;
}
// 兄弟节点必然是黑色
if (isBlack(sibling.left) && isBlack(sibling.right)) {
// 兄弟节点没有1个红色子节点，父节点要向下跟兄弟节点合并
boolean parentBlack = isBlack(parent);
black(parent);
red(sibling);
if (parentBlack) {
afterRemove(parent);
}
} else { // 兄弟节点至少有1个红色子节点，向兄弟节点借元素
// 兄弟节点的左边是黑色，兄弟要先旋转
if (isBlack(sibling.right)) {
rotateRight(sibling);
sibling = parent.right;
}
color(sibling, colorOf(parent));
black(sibling.right);
black(parent);
rotateLeft(parent);
}
} else { // 被删除的节点在右边，兄弟节点在左边
if (isRed(sibling)) { // 兄弟节点是红色
black(sibling);
red(parent);
rotateRight(parent);
// 更换兄弟
sibling = parent.left;
}
// 兄弟节点必然是黑色
if (isBlack(sibling.left) && isBlack(sibling.right)) {
// 兄弟节点没有1个红色子节点，父节点要向下跟兄弟节点合并
boolean parentBlack = isBlack(parent);
black(parent);
red(sibling);
if (parentBlack) {
afterRemove(parent);
}
} else { // 兄弟节点至少有1个红色子节点，向兄弟节点借元素
// 兄弟节点的左边是黑色，兄弟要先旋转
if (isBlack(sibling.left)) {
rotateLeft(sibling);
sibling = parent.left;
}
color(sibling, colorOf(parent));
black(sibling.left);
black(parent);
rotateRight(parent);
}
}
}
private Node<K, V> predecessor(Node<K, V> node) {
if (node == null) return null;
// 前驱节点在左子树当中（left.right.right.right....）
Node<K, V> p = node.left;
if (p != null) {
while (p.right != null) {
p = p.right;
}
return p;
}
// 从父节点、祖父节点中寻找前驱节点
while (node.parent != null && node == node.parent.left) {
node = node.parent;
}
// node.parent == null
// node == node.parent.right
return node.parent;
}
private Node<K, V> successor(Node<K, V> node) {
if (node == null) return null;
// 前驱节点在左子树当中（right.left.left.left....）
Node<K, V> p = node.right;
if (p != null) {
while (p.left != null) {
p = p.left;
}
return p;
}
// 从父节点、祖父节点中寻找前驱节点
while (node.parent != null && node == node.parent.right) {
node = node.parent;
}
return node.parent;
}
private Node<K, V> node(K key) {
Node<K, V> node = root;
while (node != null) {
int cmp = compare(key, node.key);
if (cmp == 0) return node;
if (cmp > 0) {
node = node.right;
} else { // cmp < 0
node = node.left;
}
}
return null;
}
private void afterPut(Node<K, V> node) {
Node<K, V> parent = node.parent;
// 添加的是根节点 或者 上溢到达了根节点
if (parent == null) {
black(node);
return;
}
// 如果父节点是黑色，直接返回
if (isBlack(parent)) return;
// 叔父节点
Node<K, V> uncle = parent.sibling();
// 祖父节点
Node<K, V> grand = red(parent.parent);
if (isRed(uncle)) { // 叔父节点是红色【B树节点上溢】
black(parent);
black(uncle);
// 把祖父节点当做是新添加的节点
afterPut(grand);
return;
}
// 叔父节点不是红色
if (parent.isLeftChild()) { // L
if (node.isLeftChild()) { // LL
black(parent);
} else { // LR
black(node);
rotateLeft(parent);
}
rotateRight(grand);
} else { // R
if (node.isLeftChild()) { // RL
black(node);
rotateRight(parent);
} else { // RR
black(parent);
}
rotateLeft(grand);
}
}
private void rotateLeft(Node<K, V> grand) {
Node<K, V> parent = grand.right;
Node<K, V> child = parent.left;
grand.right = child;
parent.left = grand;
afterRotate(grand, parent, child);
}
private void rotateRight(Node<K, V> grand) {
Node<K, V> parent = grand.left;
Node<K, V> child = parent.right;
grand.left = child;
parent.right = grand;
afterRotate(grand, parent, child);
}
private void afterRotate(Node<K, V> grand, Node<K, V> parent, Node<K, V> child) {
// 让parent称为子树的根节点
parent.parent = grand.parent;
if (grand.isLeftChild()) {
grand.parent.left = parent;
} else if (grand.isRightChild()) {
grand.parent.right = parent;
} else { // grand是root节点
root = parent;
}
// 更新child的parent
if (child != null) {
child.parent = grand;
}
// 更新grand的parent
grand.parent = parent;
}
private Node<K, V> color(Node<K, V> node, boolean color) {
if (node == null) return node;
node.color = color;
return node;
}
private Node<K, V> red(Node<K, V> node) {
return color(node, RED);
}
private Node<K, V> black(Node<K, V> node) {
return color(node, BLACK);
}
private boolean colorOf(Node<K, V> node) {
return node == null ? BLACK : node.color;
}
private boolean isBlack(Node<K, V> node) {
return colorOf(node) == BLACK;
}
private boolean isRed(Node<K, V> node) {
return colorOf(node) == RED;
}
private int compare(K e1, K e2) {
if (comparator != null) {
return comparator.compare(e1, e2);
}
return ((Comparable<K>)e1).compareTo(e2);
}
private void keyNotNullCheck(K key) {
if (key == null) {
throw new IllegalArgumentException("key must not be null");
}
}
private static class Node<K, V> {
K key;
V value;
boolean color = RED;
Node<K, V> left;
Node<K, V> right;
Node<K, V> parent;
public Node(K key, V value, Node<K, V> parent) {
this.key = key;
this.value = value;
this.parent = parent;
}
public boolean isLeaf() {
return left == null && right == null;
}
public boolean hasTwoChildren() {
return left != null && right != null;
}
public boolean isLeftChild() {
return parent != null && this == parent.left;
}
public boolean isRightChild() {
return parent != null && this == parent.right;
}
public Node<K, V> sibling() {
if (isLeftChild()) {
return parent.right;
}
if (isRightChild()) {
return parent.left;
}
return null;
}
}
}
static void test1() {
System.out.println("--------------------------------------- 映射 -- 红黑树实现");
Map<String, Integer> map = new TreeMap<>();
map.put("c", 2);
map.put("a", 5);
map.put("b", 6);
map.put("a", 8);
map.traversal(new Visitor<String, Integer>() {
public boolean visit(String key, Integer value) {
System.out.println(key + "_" + value);
return false;
}
});
}
// 单词量统计
static void test2() {
System.out.println("--------------------------------------- 单词量统计");
FileInfo fileInfo = Files.read("/Users/alange/eclipse-workspace/08-红黑树/src/alangeit",
new String[]{"java"});
System.out.println("文件数量：" + fileInfo.getFiles());
System.out.println("代码行数：" + fileInfo.getLines());
String[] words = fileInfo.words();
System.out.println("单词数量：" + words.length);
Map<String, Integer> map = new TreeMap<>();
for (int i = 0; i < words.length; i++) {
Integer count = map.get(words[i]);
count = (count == null) ? 1 : (count + 1);
map.put(words[i], count);
}
map.traversal(new Visitor<String, Integer>() {
public boolean visit(String key, Integer value) {
System.out.println(key + "_" + value);
return false;
}
});
}
Map 与 Set
◼Map 的所有 key 组合在一起，其实就是一个 Set
◼因此，Set 可以间接利用 Map 来作内部实现
Set.java
package alangeit.set;
public interface Set<E> {
int size();
boolean isEmpty();
void clear();
boolean contains(E element);
void add(E element);
void remove(E element);
void traversal(Visitor<E> visitor);
public static abstract class Visitor<E> {
boolean stop;
public abstract boolean visit(E element);
}
}
TreeSet.java
package alangeit.set;
import alangeit.map.Map;
import alangeit.map.TreeMap;
public class TreeSet<E> implements Set<E> {
Map<E, Object> map = new TreeMap<>();
@Override
public int size() {
return map.size();
}
@Override
public boolean isEmpty() {
return map.isEmpty();
}
@Override
public void clear() {
map.clear();
}
@Override
public boolean contains(E element) {
return map.containsKey(element);
}
@Override
public void add(E element) {
map.put(element, null);
}
@Override
public void remove(E element) {
map.remove(element);
}
@Override
public void traversal(Visitor<E> visitor) {
map.traversal(new Map.Visitor<E, Object>() {
public boolean visit(E key, Object value) {
return visitor.visit(key);
}
});
}
}
static void test() {
Set<String> set = new TreeSet<>();
set.add("c");
set.add("b");
set.add("c");
set.add("c");
set.add("a");
set.traversal(new Set.Visitor<String>() {
public boolean visit(String element) {
System.out.println(element);
return false;
}
});
}
TreeMap分析
◼ 时间复杂度(平均)
添加、删除、搜索:O(logn)
◼特点
Key 必须具备可比较性
元素的分布是有顺序的
◼ 在实际应用中，很多时候的需求
Map 中存储的元素不需要讲究顺序
Map 中的 Key 不需要具备可比较性
◼不考虑顺序、不考虑 Key 的可比较性，Map 有更好的实现方案，平均时间复杂度可以达到 O(1)
那就是采取哈希表来实现 Map
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