Map
Map 在有些编程语言中也叫字典, dictionary, Python, OC, Swift
每一个Key 有唯一性, 对应一个Value
可以使用链表和平衡二叉搜索树等结构来实现
Map 接口
public interface Map<K, V> {
int size();
boolean isEmpty();
void clear();
V put(K key, V value);
V get(K key);
V remove(K key);
boolean containsKey(K key);
boolean containsValue(V value);
void traversal(Visitor<K, V> visitor);
public static abstract class Visitor<K, V> {
boolean stop;
public abstract boolean visit(K key, V value);
}
}
具体实现
使用红黑树
key 不能为null, 且有可比较性
每个节点中存储key 和value
遍历时使用中序遍历, 按照一定顺序输出
时间复杂度, 添加删除, 搜索都是O(logn)
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;
}
@Override
public int size() {
return size;
}
@Override
public boolean isEmpty() {
return size == 0;
}
@Override
public void clear() {
root = null;
size = 0;
}
@Override
public V put(K key, V value) {
keyNotNullCheck(key);
// 只有一个根节点
if (root == null) {
root = new Node<>(key, value, null);
size++;
afterPut(root);
return null;
}
// 其他节点
// 找到父节点
Node<K, V> parent = root;
Node<K, V> node = root;
int cmp = 0;
while (node != null) {
cmp = compare(key, node.key);
parent = node;
if (cmp > 0) {
node = node.right;
} else if (cmp < 0) {
node = node.left;
} else {
node.key = key;
V oldValue = node.value;
node.value = value;
return oldValue;
}
}
Node<K, V> newNode = new Node<>(key, value, parent);
if (cmp > 0) {
parent.right = newNode;
} else {
parent.left = newNode;
}
size++;
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 (valEaquals(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 valEaquals(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
node.key = s.key;
node.value = s.value;
// 删除后继节点
node = s;
}
// 删除node 的节点, 度为1 或者0
Node<K, V> replacement = node.left != null ? node.left : node.right;
if (replacement != null) { // 度为1 的节点
// 更改parent
replacement.parent = node.parent;
if (node.parent == null) {// node 是度为1 的节点且是根节点
root = replacement;
} else if (node == node.parent.left) {
node.parent.left = replacement;
} else {
node.parent.right = replacement;
}
// 被删除的节点调整
afterRomove(replacement);
} else if (node.parent == null) {// 叶子节点, 且为根节点
root = null;
// 被删除的节点调整
afterRomove(node);
} 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;
}
// 被删除的节点调整
afterRomove(node);
}
return oldValue;
}
private void afterRomove(Node<K, V> node) {
// 如果删除的节点是红色
// if (isRed(node)) return;
// 用以取代node 的子节点是红色
if (isRed(node)) {
black(node);
return;
}
Node<K, V> parent = node.parent;
// 删除的是根节点
if (parent == null) return;
// 删除的时黑色叶子节点
// 判断被删除的node 是左还是右边
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);
sibling = parent.right;
}
// 兄弟节点必然是黑色
if (isBlack(sibling.left) && isBlack(sibling.right)) {
// 兄弟节点没有一个是红色子节点, 父节点要向下合并
boolean parentBlack = isBlack(parent);
black(parent);
red(sibling);
if (parentBlack) {
afterRomove(parent);
}
} else {// 兄弟节点至少有一个是红色子节点
// 兄弟节点的左边是黑色, 兄弟要旋转
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);
sibling = parent.left;
}
// 兄弟节点必然是黑色
if (isBlack(sibling.left) && isBlack(sibling.right)) {
// 兄弟节点没有一个是红色子节点, 父节点要向下合并
boolean parentBlack = isBlack(parent);
black(parent);
red(sibling);
if (parentBlack) {
afterRomove(parent);
}
} else {// 兄弟节点至少有一个是红色子节点
// 兄弟节点的左边是黑色, 兄弟要旋转
if (isBlack(sibling.left)) {
rotateLeft(sibling);
sibling = parent.left;
}
color(sibling, colorOf(parent));
black(sibling.left);
black(parent);
rotateRight(parent);
}
}
}
private Node<K, V> 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 {
node = node.left;
}
}
return null;
}
private void afterPut(Node<K, V> node) {
Node<K, V> parent = node.parent;
// 如果是根节点
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
black(parent);
black(uncle);
// 把祖父节点当做是新添加的节点
afterPut(red(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) {
// parent 成为子树的根节点
parent.parent = grand.parent;
if (grand.isLeftChild()) {
grand.parent.left = parent;
} else if (grand.isRightChild()) {
grand.parent.right = parent;
} else { // grand 是根节点
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 isRed(Node<K, V> node) {
return colorOf(node) == RED;
}
private boolean isBlack(Node<K, V> node) {
return colorOf(node) == BLACK;
}
private int compare(K e1, K e2) {
if (comparator != null) {
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 Node<K, V> predecessor(Node<K, V> node) {
Node<K, V> p = node.left;
// 前驱节点在左子树中, left.right.right.....
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;
}
return node.parent;
}
private Node<K, V> successor(Node<K, V> node) {
Node<K, V> p = node.right;
// 前驱节点在左子树中, right.left.left.....
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 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;
}
}
}
哈希表
有些需求为, Map 中存储的元素不需要讲究顺序, Key不需要可比较性, 平均的复杂度可以达到O(1)
可以使用哈希表来实现Map
哈希表也叫散列函数, 是时间换空间的典型应用, 内部的数组元素, 也叫桶Bucket, 整个数组叫Buckets, 或者Bucket Array
哈希冲突
也叫作哈希碰撞, 两个不同key, 经过哈希计算出相同结果
解决哈希冲突的常见方法
- 开放定址法, 按照一定规则向其他地址探测, 直到遇到空桶
- 再哈希法, 设计多个哈希函数
- 链地址法, 比如通过链表将同一个index 的元素串起来
JDK1.8 的后续冲突解决方案
- 默认使用单向链表将元素串起来, 每次从链表头部遍历, 比双向链表少一个指针, 节省内存空间
- 在添加元素时, 可能会由单向链表转为红黑树存储元素
- 当哈希表容量 >= 64, 且单向链表节点数量大于8时
- 当红黑树节点数量少到一定程度时, 又会转为单向链表
- JDK1.8 中哈希表使用链表+红黑树解决哈希冲突
哈希表
哈希函数实现步骤
- 先生成key 的哈希值, 必须为整数
- 再让key 的哈希值跟数组的大小进行相关运算, 生成一个索引值
- 可以使用&位运算取代%运算, 数组长度为2的幂次方
- 良好的哈希函数, 哈希值均匀分布, 减少哈希冲突次数, 提升哈希表性能
生成key 的哈希值
key 的常见种类, 整数, 浮点数, 字符串, 自定义对象
尽量让每个key 的哈希值是唯一的, 让key 的所有信息参与运算
在Java 中 HashMap 的key 必须实现hashCode, equals 方法, 也允许key 为null
-
整数, 整数值作为哈希值
-
浮点数, 将存储的二进制格式转为整数值
-
Long 和Double, 高32bit 和低32bit 混合计算出32bit 的哈希值, 充分利用所有信息计算出哈希值
long,double生成哈希值
- 字符串, 由若干字符组成, 例如jack, ((jn + a) * n + c) * n + k, jdk中n 为31, 31是一个奇素数, 31i, 优化为(i<<5)-i
自定义对象哈希值
public class Person {
private int age;
private float height;
private String name;
public Person(int age, float height, String name) {
this.age = age;
this.height = height;
this.name = name;
}
/**
* hash 冲突时, 比较2 个key 是否相等
*/
@Override
public boolean equals(Object obj) {
if (this == obj) return true;
if (obj == null || obj.getClass() != getClass()) return false;
//if (obj == null | !(obj instanceof Person)) return false;
Person person = (Person) obj;
return person.age == age
&& person.height == height
&& (person.name == null ? name == null : person.name.equals(name));
}
/**
* 计算索引时调用
*/
@Override
public int hashCode() {
int hashCode = Integer.hashCode(age);
hashCode = hashCode *31 + Float.hashCode(height);
hashCode = hashCode * 31 + (name != null ? name.hashCode() : 0);
return hashCode;
}
}
- 自定义对象作为key, 最好同时重新hashCode, 和equals
- equals, 用以判断两个key 是否为同一个key
- 自反性, 对于任何非null 的x, x.equals(x), 必须返回true
- 对称性, 对于任何非null 的x, y, 如果y.equals(x) 返回true, x.equals(y) 必须返回true
- 传递性, 对于任何非null 的x, y, z, 如果x.equals(y), y.equals(z) 返回true, 那么x.equals(z) 必须返回true
- 一致性, 对于任何非null 的x, y, 只要equals 的比较操作在对象中所用的信息没有被修改, 多次调用equals 就会一致返回true, 或者一致返回false
- 对于任何非null 的x, x.equals(null) 必须返回false
- hashCode, 必须保证equals 为true 的2 个key 的哈希值一样, 反过来, hashCode 相等的key, 不一定equals 的true
哈希值相等, 索引必然相等
哈希值不等, 哈希函数求结果索引, 可能相等, 取模运算可能遇到相等情况
HashMap 的key 必须实现hashCode 和equals, 也允许key 为null
完整代码
public class HashMap<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>[] table;
private static final int DEFAULT_CAPACITY = 1 << 4;
private static final float DEFAULT_LOAD_FACTOR = 0.75f;
public HashMap() {
table = new Node[DEFAULT_CAPACITY];
}
@Override
public int size() {
return size;
}
@Override
public boolean isEmpty() {
return size == 0;
}
@Override
public void clear() {
if (size == 0) return;
size = 0;
for (int i = 0; i < table.length; i++) {
table[i] = null;
}
}
@Override
public V put(K key, V value) {
resize();
int index = index(key);
// 取出index 位置的红黑树根节点
Node<K, V> root = table[index];
if (root == null) {
root = createNode(key, value, null);
table[index] = root;
size++;
fixAfterPut(root);
return null;
}
// 添加新的节点到红黑树上
Node<K, V> parent = root;
Node<K, V> node = root;
int cmp = 0;
K k1 = key;
int h1 = hash(k1);
Node<K, V> result = null;
boolean searched = false; // 是否已经搜索过这个key
do {
parent = node;
K k2 = node.key;
int h2 = node.hash;
if (h1 > h2) {
cmp = 1;
} else if (h1 < h2) {
cmp = -1;
} else if (Objects.equals(k1, k2)) {
cmp = 0;
} else if (k1 != null && k2 != null
&& k1.getClass() == k2.getClass()
&& k1 instanceof Comparable
&& (cmp = ((Comparable) k1).compareTo(k2)) != 0) {
// cmp > 0, < 0, == 0
// 为0 时, 不考虑是同一个对象, 往后扫描
} else if (searched) { // 扫描, 然后再根据内存地址大小决定左右, 已经扫描过这个key
cmp = System.identityHashCode(k1) - System.identityHashCode(k2);
} else { // searched == false, 未扫描过这个key
if ((node.left != null && (result = node(node.left, k1)) != null)
|| (node.right != null && (result = node(node.right, k1)) != null)) {
node = result;
cmp = 0;
} else {// 不存在这个key
searched = true;
cmp = System.identityHashCode(k1) - System.identityHashCode(k2);
}
}
if (cmp > 0) {
node = node.right;
} else if (cmp < 0) {
node = node.left;
} else {
V oldValue = node.value;
node.key = key;
node.value = value;
node.hash = h1;
return oldValue;
}
} while (node != null);
// 看看插入到父节点的哪个位置
Node<K, V> newNode = createNode(key, value, parent);
if (cmp > 0) {
parent.right = newNode;
} else {
parent.left = newNode;
}
size++;
fixAfterPut(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 (size == 0) return false;
Queue<Node<K, V>> queue = new LinkedList<>();
for (int i = 0; i < table.length; i++) {
if (table[i] == null) continue;
queue.offer(table[i]);
while (!queue.isEmpty()) {
Node<K, V> node = queue.poll();
if (Objects.equals(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 (size == 0 || visitor == null) return;
Queue<Node<K, V>> queue = new LinkedList<>();
for (int i = 0; i < table.length; i++) {
if (table[i] == null) continue;
queue.offer(table[i]);
while (!queue.isEmpty()) {
Node<K, V> node = queue.poll();
if (visitor.visit(node.key, node.value)) return;
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
}
}
public void print() {
if (size == 0) return;
for (int i = 0; i < table.length; i++) {
final Node<K, V> root = table[i];
System.out.println("[index = " + i + "]");
BinaryTrees.println(new BinaryTreeInfo() {
@Override
public Object string(Object node) {
return node;
}
@Override
public Object root() {
return root;
}
@Override
public Object right(Object node) {
return ((Node<K, V>)node).right;
}
@Override
public Object left(Object node) {
return ((Node<K, V>)node).left;
}
});
System.out.println("-----------------------------------");
}
}
protected Node<K, V> createNode(K key, V value, Node<K, V> parent) {
return new Node<>(key, value, parent);
}
protected void afterRemove(Node<K, V> willNode, Node<K, V> removeNode) {}
private void resize() {
// 装填因子 <= 0.75
if (size / table.length <= DEFAULT_LOAD_FACTOR) return;
Node<K, V> []oldTable = table;
table = new Node[oldTable.length << 1];
Queue<Node<K, V>> queue = new LinkedList<>();
for (int i = 0; i < oldTable.length; i++) {
if (oldTable[i] == null) continue;
queue.offer(oldTable[i]);
while (!queue.isEmpty()) {
Node<K, V> node = queue.poll();
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
// 挪动放到最后
moveNode(node);
}
}
}
private void moveNode(Node<K, V> newNode) {
newNode.parent = null;
newNode.left = null;
newNode.right = null;
newNode.color = RED;
int index = index(newNode);
// 取出index 位置的红黑树根节点
Node<K, V> root = table[index];
if (root == null) {
root = newNode;
table[index] = root;
fixAfterPut(root);
return;
}
// 添加新的节点到红黑树上
Node<K, V> parent = root;
Node<K, V> node = root;
int cmp = 0;
K k1 = newNode.key;
int h1 = newNode.hash;
do {
parent = node;
K k2 = node.key;
int h2 = node.hash;
if (h1 > h2) {
cmp = 1;
} else if (h1 < h2) {
cmp = -1;
} else if (k1 != null && k2 != null
&& k1.getClass() == k2.getClass()
&& k1 instanceof Comparable
&& (cmp = ((Comparable) k1).compareTo(k2)) != 0) {
// cmp > 0, < 0, == 0
// 为0 时, 不考虑是同一个对象, 往后扫描
} else {
cmp = System.identityHashCode(k1) - System.identityHashCode(k2);
}
if (cmp > 0) {
node = node.right;
} else if (cmp < 0) {
node = node.left;
}
} while (node != null);
// 看看插入到父节点的哪个位置
newNode.parent = parent;
if (cmp > 0) {
parent.right = newNode;
} else {
parent.left = newNode;
}
fixAfterPut(newNode);
}
protected V remove(Node<K, V> node) {
if (node == null) return null;
size--;
Node<K, V> willNode = node;
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
node.key = s.key;
node.value = s.value;
node.hash = s.hash;
// 删除后继节点
node = s;
}
// 删除node 的节点, 度为1 或者0
Node<K, V> replacement = node.left != null ? node.left : node.right;
int index = index(node);
if (replacement != null) { // 度为1 的节点
// 更改parent
replacement.parent = node.parent;
if (node.parent == null) {// node 是度为1 的节点且是根节点
table[index] = replacement;
} else if (node == node.parent.left) {
node.parent.left = replacement;
} else {
node.parent.right = replacement;
}
// 被删除的节点调整
fixAfterRomove(replacement);
} else if (node.parent == null) {// 叶子节点, 且为根节点
table[index] = null;
// 被删除的节点调整
// afterRomove(node);
} 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;
}
// 被删除的节点调整
fixAfterRomove(node);
}
afterRemove(willNode, node);
return oldValue;
}
private Node<K, V> successor(Node<K, V> node) {
if (node == null) return null;
Node<K, V> p = node.right;
// 前驱节点在左子树中, right.left.left.....
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> root = table[index(key)];
return root == null ? null : node(root, key);
}
private Node< K, V> node(Node<K, V> node, K k1) {
int h1 = hash(k1);
// 存储查找结果
Node<K, V> result = null;
int cmp = 0;
while (node != null) {
K k2 = node.key;
int h2 = node.hash;
// 先比较hash 值
if (h1 > h2) {
node = node.right;
} else if (h1 < h2) {
node = node.left;
} else if (Objects.equals(k1, k2)) {
return node;
} else if (k1 != null && k2 != null
&& k1.getClass() == k2.getClass()
&& k1 instanceof Comparable
&& (cmp = ((Comparable) k1).compareTo(k2)) != 0) {
node = cmp > 0 ? node.right : node.left;
// if (cmp > 0) {
// node = node.right;
// } else if (cmp < 0) {
// node = node.left;
// } else {
// return node;
// }
// hash 值相等, 不具备可比较性, 不是equals
} else if (node.right != null && (result = node(node.right, k1)) != null) {
return result;
} else { // 左边查找
node = node.left;
}
// else if (node.left != null && (result = node(node.left, k1)) != null) {
// return result;
// } else {
// return null;
// }
// int cmp = compare(key, node.key, h1, node.hash);
// if (cmp == 0) return node;
// if (cmp > 0) {
// node = node.right;
// } else if (cmp < 0) {
// node = node.left;
// }
}
return null;
}
/**
* 根据key 生成对应的索引, 在桶数组中的位置
* @param key
* @return
*/
private int index(K key) {
return hash(key) & (table.length - 1);
}
private int hash(K key) {
if (key == null) return 0;
int hash = key.hashCode();
return hash ^ (hash >>> 16);
}
private int index(Node<K, V> node) {
return node.hash & (table.length - 1);
}
private void fixAfterRomove(Node<K, V> node) {
// 如果删除的节点是红色
// if (isRed(node)) return;
// 用以取代node 的子节点是红色
if (isRed(node)) {
black(node);
return;
}
Node<K, V> parent = node.parent;
// 删除的是根节点
if (parent == null) return;
// 删除的时黑色叶子节点
// 判断被删除的node 是左还是右边
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);
sibling = parent.right;
}
// 兄弟节点必然是黑色
if (isBlack(sibling.left) && isBlack(sibling.right)) {
// 兄弟节点没有一个是红色子节点, 父节点要向下合并
boolean parentBlack = isBlack(parent);
black(parent);
red(sibling);
if (parentBlack) {
fixAfterRomove(parent);
}
} else {// 兄弟节点至少有一个是红色子节点
// 兄弟节点的左边是黑色, 兄弟要旋转
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);
sibling = parent.left;
}
// 兄弟节点必然是黑色
if (isBlack(sibling.left) && isBlack(sibling.right)) {
// 兄弟节点没有一个是红色子节点, 父节点要向下合并
boolean parentBlack = isBlack(parent);
black(parent);
red(sibling);
if (parentBlack) {
fixAfterRomove(parent);
}
} else {// 兄弟节点至少有一个是红色子节点
// 兄弟节点的左边是黑色, 兄弟要旋转
if (isBlack(sibling.left)) {
rotateLeft(sibling);
sibling = parent.left;
}
color(sibling, colorOf(parent));
black(sibling.left);
black(parent);
rotateRight(parent);
}
}
}
private void fixAfterPut(Node<K, V> node) {
Node<K, V> parent = node.parent;
// 如果是根节点
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
black(parent);
black(uncle);
// 把祖父节点当做是新添加的节点
fixAfterPut(red(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) {
// parent 成为子树的根节点
parent.parent = grand.parent;
if (grand.isLeftChild()) {
grand.parent.left = parent;
} else if (grand.isRightChild()) {
grand.parent.right = parent;
} else { // grand 是根节点
table[index(grand)] = 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 isRed(Node<K, V> node) {
return colorOf(node) == RED;
}
private boolean isBlack(Node<K, V> node) {
return colorOf(node) == BLACK;
}
protected static class Node<K, V> {
int hash;
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;
int hash = key == null ? 0 : key.hashCode();
this.hash = hash ^ (hash >>> 16);
this.value = value;
this.parent = parent;
}
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;
}
@Override
public String toString() {
return "Node_" + key + "_" + value;
}
}
}
HashMap 升级为LinkedHashMap
在HashMap 的基础上维护元素的添加顺序, 使得遍历结果遵从添加顺序
删除度为2 的节点31, 调整顺序, 更换node 的前驱后继节点的连接位置
删除数据调整顺序
public class LinkedHashMap<K, V> extends HashMap<K, V> {
private LinkedNode<K, V> first;
private LinkedNode<K, V> last;
@Override
public void clear() {
super.clear();
first = null;
last = null;
}
@Override
public boolean containsValue(V value) {
LinkedNode<K, V> node = first;
while (node != null) {
if (Objects.equals(value, node.value)) return true;
node = node.next;
}
return false;
}
@Override
public void traversal(Visitor<K, V> visitor) {
if (visitor == null) return;
LinkedNode<K, V> node = first;
while (node != null) {
if (visitor.visit(node.key, node.value)) return;
node = node.next;
}
}
@Override
protected void afterRemove(Node<K, V> willNode, Node<K, V> removeNode) {
LinkedNode<K, V> node1 = (LinkedNode<K, V>) willNode;
LinkedNode<K, V> node2 = (LinkedNode<K, V>) removeNode;
if (node1 != node2) {
// 交换两者在链表中的位置
// 交换prev
LinkedNode<K, V> tmp = node1.prev;
node1.prev = node2.prev;
node2.prev = tmp;
if (node1.prev == null) {
first = node1;
} else {
node1.prev.next = node1;
}
if (node2.prev == null) {
first = node2;
} else {
node2.prev.next = node2;
}
// 交换next
tmp = node1.next;
node1.next = node2.next;
node2.next = tmp;
if (node1.next == null) {
last = node1;
} else {
node1.next.prev = node1;
}
if (node2.next == null) {
last = node2;
} else {
node2.next.prev = node2;
}
}
LinkedNode<K, V> prev = node2.prev;
LinkedNode<K, V> next = node2.next;
if (prev == null) {
first = next;
} else {
prev.next = next;
}
if (next == null) {
last = prev;
} else {
next.prev = prev;
}
}
@Override
protected Node<K, V> createNode(K key, V value, Node<K, V> parent) {
LinkedNode<K, V> node = new LinkedNode(key, value, parent);
if (first == null) {
first = last = node;
} else {
last.next = node;
node.prev = last;
last = node;
}
return node;
}
private static class LinkedNode<K, V> extends Node<K, V> {
LinkedNode<K, V> prev;
LinkedNode<K, V> next;
public LinkedNode(K key, V value, Node<K, V> parent) {
super(key, value, parent);
}
}
}
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