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二十、映射(Map)-实现TreeMap、TreeSet

二十、映射(Map)-实现TreeMap、TreeSet

作者: 咸鱼Jay | 来源:发表于2022-02-04 09:02 被阅读0次

    1、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)


    映射(Map)

    Map 的每一个 key 是唯一的

    2、Map的接口设计

    Map的接口设计

    3、TreeMap实现

    类似SetMap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

    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);
            
            // 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;
        }
    
        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);
            }
        }
    
        @Override
        public V get(K key) {
            Node<K, V> node = findNode(key);
            return node != null ? node.value : null;
        }
    
        @Override
        public V remove(K key) {
            return remove(findNode(key));
        }
    
        @Override
        public boolean containsKey(K key) {
            return findNode(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;
        }
        
        private boolean valEquals(V v1, V v2) {
            return v1 == null ? v2 == null : v1.equals(v2);
        }
        
        @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 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> 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> findNode(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 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;
            }
        }
    }
    

    4、Map 与 Set

    Map的所有key组合在一起,其实就是一个Set
    因此,Set还可以间接利用Map来作内部实现

    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);
                }
            });
        }
    }
    

    代码链接

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