有序表
- 搜索二叉树
- 搜索二叉树查询key（查询某个key存在还是不存在）
- 搜索二叉树插入新的key
- 搜索二叉树删除key
- 搜索二叉树特别不讲究
- AVL树、SB树、红黑树的共性
- AVL树、SB树、红黑树的不同
- AVL树
- AVL数
- SB树（size-balance-tree）
- SB树在使用时候的改进
- SB树代码
- 跳表（skiplist）
- 跳表代码
- 红黑树
- 习题1 给定一个数组arr，和两个整数a和b（a<=b）求arr中有多少个子数组，累加和在[a,b]这个范围上返回达标的子数组数量
- 习题2 有一个滑动窗口（讲过的）：1）L是滑动窗口最左位置、R是滑动窗口最右位置，一开始LR都在数组左侧 2）任何一步都可能R往右动，表示某个数进了窗口 3）任何一步都可能L往右动，表示某个数出了窗口 想知道每一个窗口状态的中位数
- 习题3 设计一个结构包含如下两个方法：void add(int index, int num)：把num加入到index位置 int get(int index) ：取出index位置的值 void remove(int index) ：把index位置上的值删除 要求三个方法时间复杂度O(logN)
- 改写有序表的题目核心点
搜索二叉树
搜索二叉树一定要说明以什么标准来排序
经典的搜索二叉树，树上没有重复的用来排序的key值
如果有重复节点的需求，可以在一个节点内部增加数据项
搜索二叉树查询key（查询某个key存在还是不存在）
1）如果当前节点的value==key，返回true
2）如果当前节点的value<key，当前节点向左移动
3）如果当前节点的value>key，当前节点向右移动
4）如果当前节点变成null，返回false
搜索二叉树插入新的key
和查询过程一样，但当前节点滑到空的时候，就插入在这里
搜索二叉树删除key
1）先找到key所在的节点
2）如果该节点没有左孩子、没有右孩子，直接删除即可
3）如果该节点有左孩子、没有右孩子，直接用左孩子顶替该节点
4）如果该节点没有左孩子、有右孩子，直接用右孩子顶替该节点
5）如果该节点有左孩子、有右孩子，用该节点后继节点顶替该节点
搜索二叉树特别不讲究
1）基础的搜索二叉树，添加、删除时候不照顾平衡性
2）数据状况很差时，性能就很差
给搜索二叉树引入两个动作：左旋、右旋
AVL树、SB树、红黑树的共性
1）都是搜索二叉树
2）插入、删除、查询（一切查询）搜索二叉树怎么做，这些结构都这么做
3）使用调整的基本动作都只有左旋、右旋
4）插入、删除时，从最底层被影响到的节点开始，对往上路径的节点做平衡性检查
5）因为只对一条向上路径的每个节点做O(1)的检查和调整，所以可以做到O(logN)
AVL树、SB树、红黑树的不同
1）平衡性的约束不同
AVL树最严格、SB树稍宽松、红黑树最宽松
2）插入、删除和搜索二叉树一样，但是额外，做各自的平衡性调整。各自的平衡性调整所使用的动作都是左旋或者右旋
AVL树
1）最严格的平衡性，任何节点左树高度和右树高度差不超过1
2）往上沿途检查每个节点时，都去检查四种违规情况：LL、RR、LR、RL
3）不同情况虽然看起来复杂，但是核心点是：
LL（做一次右旋）、RR（做一次左旋）
LR和RL（利用旋转让底层那个上到顶部）
当即为LL，也为LR时，用LL的方法
当既为RR，也为RL时，用RR的方法
AVL数
public static class AVLNode<K extends Comparable<K>, V> {
public K k;
public V v;
public AVLNode<K, V> l;
public AVLNode<K, V> r;
public int h;
public AVLNode(K key, V value) {
k = key;
v = value;
h = 1;
}
}
public static class AVLTreeMap<K extends Comparable<K>, V> {
private AVLNode<K, V> root;
private int size;
public AVLTreeMap() {
root = null;
size = 0;
}
private AVLNode<K, V> rightRotate(AVLNode<K, V> cur) {
AVLNode<K, V> left = cur.l;
cur.l = left.r;
left.r = cur;
cur.h = Math.max((cur.l != null ? cur.l.h : 0), (cur.r != null ? cur.r.h : 0)) + 1;
left.h = Math.max((left.l != null ? left.l.h : 0), (left.r != null ? left.r.h : 0)) + 1;
return left;
}
private AVLNode<K, V> leftRotate(AVLNode<K, V> cur) {
AVLNode<K, V> right = cur.r;
cur.r = right.l;
right.l = cur;
cur.h = Math.max((cur.l != null ? cur.l.h : 0), (cur.r != null ? cur.r.h : 0)) + 1;
right.h = Math.max((right.l != null ? right.l.h : 0), (right.r != null ? right.r.h : 0)) + 1;
return right;
}
private AVLNode<K, V> maintain(AVLNode<K, V> cur) {
if (cur == null) {
return null;
}
int leftHeight = cur.l != null ? cur.l.h : 0;
int rightHeight = cur.r != null ? cur.r.h : 0;
if (Math.abs(leftHeight - rightHeight) > 1) {
if (leftHeight > rightHeight) {
int leftLeftHeight = cur.l != null && cur.l.l != null ? cur.l.l.h : 0;
int leftRightHeight = cur.l != null && cur.l.r != null ? cur.l.r.h : 0;
if (leftLeftHeight >= leftRightHeight) {
cur = rightRotate(cur);
} else {
cur.l = leftRotate(cur.l);
cur = rightRotate(cur);
}
} else {
int rightLeftHeight = cur.r != null && cur.r.l != null ? cur.r.l.h : 0;
int rightRightHeight = cur.r != null && cur.r.r != null ? cur.r.r.h : 0;
if (rightRightHeight >= rightLeftHeight) {
cur = leftRotate(cur);
} else {
cur.r = rightRotate(cur.r);
cur = leftRotate(cur);
}
}
}
return cur;
}
private AVLNode<K, V> findLastIndex(K key) {
AVLNode<K, V> pre = root;
AVLNode<K, V> cur = root;
while (cur != null) {
pre = cur;
if (key.compareTo(cur.k) == 0) {
break;
} else if (key.compareTo(cur.k) < 0) {
cur = cur.l;
} else {
cur = cur.r;
}
}
return pre;
}
private AVLNode<K, V> findLastNoSmallIndex(K key) {
AVLNode<K, V> ans = null;
AVLNode<K, V> cur = root;
while (cur != null) {
if (key.compareTo(cur.k) == 0) {
ans = cur;
break;
} else if (key.compareTo(cur.k) < 0) {
ans = cur;
cur = cur.l;
} else {
cur = cur.r;
}
}
return ans;
}
private AVLNode<K, V> findLastNoBigIndex(K key) {
AVLNode<K, V> ans = null;
AVLNode<K, V> cur = root;
while (cur != null) {
if (key.compareTo(cur.k) == 0) {
ans = cur;
break;
} else if (key.compareTo(cur.k) < 0) {
cur = cur.l;
} else {
ans = cur;
cur = cur.r;
}
}
return ans;
}
private AVLNode<K, V> add(AVLNode<K, V> cur, K key, V value) {
if (cur == null) {
return new AVLNode<K, V>(key, value);
} else {
if (key.compareTo(cur.k) < 0) {
cur.l = add(cur.l, key, value);
} else {
cur.r = add(cur.r, key, value);
}
cur.h = Math.max(cur.l != null ? cur.l.h : 0, cur.r != null ? cur.r.h : 0) + 1;
return maintain(cur);
}
}
// 在cur这棵树上，删掉key所代表的节点
// 返回cur这棵树的新头部
private AVLNode<K, V> delete(AVLNode<K, V> cur, K key) {
if (key.compareTo(cur.k) > 0) {
cur.r = delete(cur.r, key);
} else if (key.compareTo(cur.k) < 0) {
cur.l = delete(cur.l, key);
} else {
if (cur.l == null && cur.r == null) {
cur = null;
} else if (cur.l == null && cur.r != null) {
cur = cur.r;
} else if (cur.l != null && cur.r == null) {
cur = cur.l;
} else {
AVLNode<K, V> des = cur.r;
while (des.l != null) {
des = des.l;
}
cur.r = delete(cur.r, des.k);
des.l = cur.l;
des.r = cur.r;
cur = des;
}
}
if (cur != null) {
cur.h = Math.max(cur.l != null ? cur.l.h : 0, cur.r != null ? cur.r.h : 0) + 1;
}
return maintain(cur);
}
public int size() {
return size;
}
public boolean containsKey(K key) {
if (key == null) {
return false;
}
AVLNode<K, V> lastNode = findLastIndex(key);
return lastNode != null && key.compareTo(lastNode.k) == 0 ? true : false;
}
public void put(K key, V value) {
if (key == null) {
return;
}
AVLNode<K, V> lastNode = findLastIndex(key);
if (lastNode != null && key.compareTo(lastNode.k) == 0) {
lastNode.v = value;
} else {
size++;
root = add(root, key, value);
}
}
public void remove(K key) {
if (key == null) {
return;
}
if (containsKey(key)) {
size--;
root = delete(root, key);
}
}
public V get(K key) {
if (key == null) {
return null;
}
AVLNode<K, V> lastNode = findLastIndex(key);
if (lastNode != null && key.compareTo(lastNode.k) == 0) {
return lastNode.v;
}
return null;
}
public K firstKey() {
if (root == null) {
return null;
}
AVLNode<K, V> cur = root;
while (cur.l != null) {
cur = cur.l;
}
return cur.k;
}
public K lastKey() {
if (root == null) {
return null;
}
AVLNode<K, V> cur = root;
while (cur.r != null) {
cur = cur.r;
}
return cur.k;
}
public K floorKey(K key) {
if (key == null) {
return null;
}
AVLNode<K, V> lastNoBigNode = findLastNoBigIndex(key);
return lastNoBigNode == null ? null : lastNoBigNode.k;
}
public K ceilingKey(K key) {
if (key == null) {
return null;
}
AVLNode<K, V> lastNoSmallNode = findLastNoSmallIndex(key);
return lastNoSmallNode == null ? null : lastNoSmallNode.k;
}
}
SB树（size-balance-tree）
1）让每一个叔叔节点为头的数，节点个数都不少于其任何一个侄子节点
2）也是从底层被影响节点开始向上做路径每个节点检查
3）与AVL树非常像，也是四种违规类型：LL、RR、LR、RL
4）与AVL树非常像，核心点是：
LL（做一次右旋）、RR（做一次左旋）
LR和RL（利用旋转让底层那个上到顶部）
5）与AVL树不同的是，每轮经过调整后，谁的孩子发生变化了，谁就再查
SB树在使用时候的改进
1）删除时候可以不用检查
2）就把平衡性的调整放在插入的时候
3）因为这种只要变就递归的特性，别的树没有
4）可以在节点上封装别的数据项，来增加功能
SB树代码
public static class SBTNode<K extends Comparable<K>, V> {
public K key;
public V value;
public SBTNode<K, V> l;
public SBTNode<K, V> r;
public int size; // 不同的key的数量
public SBTNode(K key, V value) {
this.key = key;
this.value = value;
size = 1;
}
}
public static class SizeBalancedTreeMap<K extends Comparable<K>, V> {
private SBTNode<K, V> root;
private SBTNode<K, V> rightRotate(SBTNode<K, V> cur) {
SBTNode<K, V> leftNode = cur.l;
cur.l = leftNode.r;
leftNode.r = cur;
leftNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
return leftNode;
}
private SBTNode<K, V> leftRotate(SBTNode<K, V> cur) {
SBTNode<K, V> rightNode = cur.r;
cur.r = rightNode.l;
rightNode.l = cur;
rightNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
return rightNode;
}
private SBTNode<K, V> maintain(SBTNode<K, V> cur) {
if (cur == null) {
return null;
}
int leftSize = cur.l != null ? cur.l.size : 0;
int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0;
int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0;
int rightSize = cur.r != null ? cur.r.size : 0;
int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0;
int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0;
if (leftLeftSize > rightSize) {
cur = rightRotate(cur);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (leftRightSize > rightSize) {
cur.l = leftRotate(cur.l);
cur = rightRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (rightRightSize > leftSize) {
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur = maintain(cur);
} else if (rightLeftSize > leftSize) {
cur.r = rightRotate(cur.r);
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
}
return cur;
}
private SBTNode<K, V> findLastIndex(K key) {
SBTNode<K, V> pre = root;
SBTNode<K, V> cur = root;
while (cur != null) {
pre = cur;
if (key.compareTo(cur.key) == 0) {
break;
} else if (key.compareTo(cur.key) < 0) {
cur = cur.l;
} else {
cur = cur.r;
}
}
return pre;
}
private SBTNode<K, V> findLastNoSmallIndex(K key) {
SBTNode<K, V> ans = null;
SBTNode<K, V> cur = root;
while (cur != null) {
if (key.compareTo(cur.key) == 0) {
ans = cur;
break;
} else if (key.compareTo(cur.key) < 0) {
ans = cur;
cur = cur.l;
} else {
cur = cur.r;
}
}
return ans;
}
private SBTNode<K, V> findLastNoBigIndex(K key) {
SBTNode<K, V> ans = null;
SBTNode<K, V> cur = root;
while (cur != null) {
if (key.compareTo(cur.key) == 0) {
ans = cur;
break;
} else if (key.compareTo(cur.key) < 0) {
cur = cur.l;
} else {
ans = cur;
cur = cur.r;
}
}
return ans;
}
// 现在，以cur为头的树上，新增，加(key, value)这样的记录
// 加完之后，会对cur做检查，该调整调整
// 返回，调整完之后，整棵树的新头部
private SBTNode<K, V> add(SBTNode<K, V> cur, K key, V value) {
if (cur == null) {
return new SBTNode<K, V>(key, value);
} else {
cur.size++;
if (key.compareTo(cur.key) < 0) {
cur.l = add(cur.l, key, value);
} else {
cur.r = add(cur.r, key, value);
}
return maintain(cur);
}
}
// 在cur这棵树上，删掉key所代表的节点
// 返回cur这棵树的新头部
private SBTNode<K, V> delete(SBTNode<K, V> cur, K key) {
cur.size--;
if (key.compareTo(cur.key) > 0) {
cur.r = delete(cur.r, key);
} else if (key.compareTo(cur.key) < 0) {
cur.l = delete(cur.l, key);
} else { // 当前要删掉cur
if (cur.l == null && cur.r == null) {
// free cur memory -> C++
cur = null;
} else if (cur.l == null && cur.r != null) {
// free cur memory -> C++
cur = cur.r;
} else if (cur.l != null && cur.r == null) {
// free cur memory -> C++
cur = cur.l;
} else { // 有左有右
SBTNode<K, V> pre = null;
SBTNode<K, V> des = cur.r;
des.size--;
while (des.l != null) {
pre = des;
des = des.l;
des.size--;
}
if (pre != null) {
pre.l = des.r;
des.r = cur.r;
}
des.l = cur.l;
des.size = des.l.size + (des.r == null ? 0 : des.r.size) + 1;
// free cur memory -> C++
cur = des;
}
}
// cur = maintain(cur);
return cur;
}
private SBTNode<K, V> getIndex(SBTNode<K, V> cur, int kth) {
if (kth == (cur.l != null ? cur.l.size : 0) + 1) {
return cur;
} else if (kth <= (cur.l != null ? cur.l.size : 0)) {
return getIndex(cur.l, kth);
} else {
return getIndex(cur.r, kth - (cur.l != null ? cur.l.size : 0) - 1);
}
}
public int size() {
return root == null ? 0 : root.size;
}
public boolean containsKey(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
SBTNode<K, V> lastNode = findLastIndex(key);
return lastNode != null && key.compareTo(lastNode.key) == 0 ? true : false;
}
// （key，value） put -> 有序表 新增、改value
public void put(K key, V value) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
SBTNode<K, V> lastNode = findLastIndex(key);
if (lastNode != null && key.compareTo(lastNode.key) == 0) {
lastNode.value = value;
} else {
root = add(root, key, value);
}
}
public void remove(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
if (containsKey(key)) {
root = delete(root, key);
}
}
public K getIndexKey(int index) {
if (index < 0 || index >= this.size()) {
throw new RuntimeException("invalid parameter.");
}
return getIndex(root, index + 1).key;
}
public V getIndexValue(int index) {
if (index < 0 || index >= this.size()) {
throw new RuntimeException("invalid parameter.");
}
return getIndex(root, index + 1).value;
}
public V get(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
SBTNode<K, V> lastNode = findLastIndex(key);
if (lastNode != null && key.compareTo(lastNode.key) == 0) {
return lastNode.value;
} else {
return null;
}
}
public K firstKey() {
if (root == null) {
return null;
}
SBTNode<K, V> cur = root;
while (cur.l != null) {
cur = cur.l;
}
return cur.key;
}
public K lastKey() {
if (root == null) {
return null;
}
SBTNode<K, V> cur = root;
while (cur.r != null) {
cur = cur.r;
}
return cur.key;
}
public K floorKey(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
SBTNode<K, V> lastNoBigNode = findLastNoBigIndex(key);
return lastNoBigNode == null ? null : lastNoBigNode.key;
}
public K ceilingKey(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
SBTNode<K, V> lastNoSmallNode = findLastNoSmallIndex(key);
return lastNoSmallNode == null ? null : lastNoSmallNode.key;
}
}
// for test
public static void printAll(SBTNode<String, Integer> head) {
System.out.println("Binary Tree:");
printInOrder(head, 0, "H", 17);
System.out.println();
}
// for test
public static void printInOrder(SBTNode<String, Integer> head, int height, String to, int len) {
if (head == null) {
return;
}
printInOrder(head.r, height + 1, "v", len);
String val = to + "(" + head.key + "," + head.value + ")" + to;
int lenM = val.length();
int lenL = (len - lenM) / 2;
int lenR = len - lenM - lenL;
val = getSpace(lenL) + val + getSpace(lenR);
System.out.println(getSpace(height * len) + val);
printInOrder(head.l, height + 1, "^", len);
}
// for test
public static String getSpace(int num) {
String space = " ";
StringBuffer buf = new StringBuffer("");
for (int i = 0; i < num; i++) {
buf.append(space);
}
return buf.toString();
}
public static void main(String[] args) {
SizeBalancedTreeMap<String, Integer> sbt = new SizeBalancedTreeMap<String, Integer>();
sbt.put("d", 4);
sbt.put("c", 3);
sbt.put("a", 1);
sbt.put("b", 2);
// sbt.put("e", 5);
sbt.put("g", 7);
sbt.put("f", 6);
sbt.put("h", 8);
sbt.put("i", 9);
sbt.put("a", 111);
System.out.println(sbt.get("a"));
sbt.put("a", 1);
System.out.println(sbt.get("a"));
for (int i = 0; i < sbt.size(); i++) {
System.out.println(sbt.getIndexKey(i) + " , " + sbt.getIndexValue(i));
}
printAll(sbt.root);
System.out.println(sbt.firstKey());
System.out.println(sbt.lastKey());
System.out.println(sbt.floorKey("g"));
System.out.println(sbt.ceilingKey("g"));
System.out.println(sbt.floorKey("e"));
System.out.println(sbt.ceilingKey("e"));
System.out.println(sbt.floorKey(""));
System.out.println(sbt.ceilingKey(""));
System.out.println(sbt.floorKey("j"));
System.out.println(sbt.ceilingKey("j"));
sbt.remove("d");
printAll(sbt.root);
sbt.remove("f");
printAll(sbt.root);
}
跳表（skiplist）
1）结构上根本和搜索二叉树无关
2）利用随机概率分布来使得高层索引可以无视数据规律，做到整体性能优良
3）思想是所有有序表中最先进的
4）结构简单就是多级单链表
跳表代码
// 跳表的节点定义
public static class SkipListNode<K extends Comparable<K>, V> {
public K key;
public V val;
public ArrayList<SkipListNode<K, V>> nextNodes;
public SkipListNode(K k, V v) {
key = k;
val = v;
nextNodes = new ArrayList<SkipListNode<K, V>>();
}
// 遍历的时候，如果是往右遍历到的null(next == null), 遍历结束
// 头(null), 头节点的null，认为最小
// node -> 头，node(null, "") node.isKeyLess(!null) true
// node里面的key是否比otherKey小，true，不是false
public boolean isKeyLess(K otherKey) {
// otherKey == null -> false
return otherKey != null && (key == null || key.compareTo(otherKey) < 0);
}
public boolean isKeyEqual(K otherKey) {
return (key == null && otherKey == null)
|| (key != null && otherKey != null && key.compareTo(otherKey) == 0);
}
}
public static class SkipListMap<K extends Comparable<K>, V> {
private static final double PROBABILITY = 0.5; // < 0.5 继续做，>=0.5 停
private SkipListNode<K, V> head;
private int size;
private int maxLevel;
public SkipListMap() {
head = new SkipListNode<K, V>(null, null);
head.nextNodes.add(null); // 0
size = 0;
maxLevel = 0;
}
// 从最高层开始，一路找下去，
// 最终，找到第0层的<key的最右的节点
private SkipListNode<K, V> mostRightLessNodeInTree(K key) {
if (key == null) {
return null;
}
int level = maxLevel;
SkipListNode<K, V> cur = head;
while (level >= 0) { // 从上层跳下层
// cur level -> level-1
cur = mostRightLessNodeInLevel(key, cur, level--);
}
return cur;
}
// 在level层里，如何往右移动
// 现在来到的节点是cur，来到了cur的level层，在level层上，找到<key最后一个节点并返回
private SkipListNode<K, V> mostRightLessNodeInLevel(K key,
SkipListNode<K, V> cur,
int level) {
SkipListNode<K, V> next = cur.nextNodes.get(level);
while (next != null && next.isKeyLess(key)) {
cur = next;
next = cur.nextNodes.get(level);
}
return cur;
}
public boolean containsKey(K key) {
if (key == null) {
return false;
}
SkipListNode<K, V> less = mostRightLessNodeInTree(key);
SkipListNode<K, V> next = less.nextNodes.get(0);
return next != null && next.isKeyEqual(key);
}
// 新增、改value
public void put(K key, V value) {
if (key == null) {
return;
}
// 0层上，最右一个，< key 的Node -> >key
SkipListNode<K, V> less = mostRightLessNodeInTree(key);
SkipListNode<K, V> find = less.nextNodes.get(0);
if (find != null && find.isKeyEqual(key)) {
find.val = value;
} else { // find == null 8 7 9
size++;
int newNodeLevel = 0;
while (Math.random() < PROBABILITY) {
newNodeLevel++;
}
// newNodeLevel
while (newNodeLevel > maxLevel) {
head.nextNodes.add(null);
maxLevel++;
}
SkipListNode<K, V> newNode = new SkipListNode<K, V>(key, value);
for (int i = 0; i <= newNodeLevel; i++) {
newNode.nextNodes.add(null);
}
int level = maxLevel;
SkipListNode<K, V> pre = head;
while (level >= 0) {
// level 层中，找到最右的 < key 的节点
pre = mostRightLessNodeInLevel(key, pre, level);
if (level <= newNodeLevel) {
newNode.nextNodes.set(level, pre.nextNodes.get(level));
pre.nextNodes.set(level, newNode);
}
level--;
}
}
}
public V get(K key) {
if (key == null) {
return null;
}
SkipListNode<K, V> less = mostRightLessNodeInTree(key);
SkipListNode<K, V> next = less.nextNodes.get(0);
return next != null && next.isKeyEqual(key) ? next.val : null;
}
public void remove(K key) {
if (containsKey(key)) {
size--;
int level = maxLevel;
SkipListNode<K, V> pre = head;
while (level >= 0) {
pre = mostRightLessNodeInLevel(key, pre, level);
SkipListNode<K, V> next = pre.nextNodes.get(level);
// 1）在这一层中，pre下一个就是key
// 2）在这一层中，pre的下一个key是>要删除key
if (next != null && next.isKeyEqual(key)) {
// free delete node memory -> C++
// level : pre -> next(key) -> ...
pre.nextNodes.set(level, next.nextNodes.get(level));
}
// 在level层只有一个节点了，就是默认节点head
if (level != 0 && pre == head && pre.nextNodes.get(level) == null) {
head.nextNodes.remove(level);
maxLevel--;
}
level--;
}
}
}
public K firstKey() {
return head.nextNodes.get(0) != null ? head.nextNodes.get(0).key : null;
}
public K lastKey() {
int level = maxLevel;
SkipListNode<K, V> cur = head;
while (level >= 0) {
SkipListNode<K, V> next = cur.nextNodes.get(level);
while (next != null) {
cur = next;
next = cur.nextNodes.get(level);
}
level--;
}
return cur.key;
}
public K ceilingKey(K key) {
if (key == null) {
return null;
}
SkipListNode<K, V> less = mostRightLessNodeInTree(key);
SkipListNode<K, V> next = less.nextNodes.get(0);
return next != null ? next.key : null;
}
public K floorKey(K key) {
if (key == null) {
return null;
}
SkipListNode<K, V> less = mostRightLessNodeInTree(key);
SkipListNode<K, V> next = less.nextNodes.get(0);
return next != null && next.isKeyEqual(key) ? next.key : less.key;
}
public int size() {
return size;
}
}
// for test
public static void printAll(SkipListMap<String, String> obj) {
for (int i = obj.maxLevel; i >= 0; i--) {
System.out.print("Level " + i + " : ");
SkipListNode<String, String> cur = obj.head;
while (cur.nextNodes.get(i) != null) {
SkipListNode<String, String> next = cur.nextNodes.get(i);
System.out.print("(" + next.key + " , " + next.val + ") ");
cur = next;
}
System.out.println();
}
}
public static void main(String[] args) {
SkipListMap<String, String> test = new SkipListMap<>();
printAll(test);
System.out.println("======================");
test.put("A", "10");
printAll(test);
System.out.println("======================");
test.remove("A");
printAll(test);
System.out.println("======================");
test.put("E", "E");
test.put("B", "B");
test.put("A", "A");
test.put("F", "F");
test.put("C", "C");
test.put("D", "D");
printAll(test);
System.out.println("======================");
System.out.println(test.containsKey("B"));
System.out.println(test.containsKey("Z"));
System.out.println(test.firstKey());
System.out.println(test.lastKey());
System.out.println(test.floorKey("D"));
System.out.println(test.ceilingKey("D"));
System.out.println("======================");
test.remove("D");
printAll(test);
System.out.println("======================");
System.out.println(test.floorKey("D"));
System.out.println(test.ceilingKey("D"));
}
红黑树
特点：每个支数的黑节点数相同，红节点不能相连，所以最长的支数的节点数比最短支数的节点数相差在两倍以内
1）平衡性规定非常诡异
2）平衡性调整最为复杂
3）优点在于每次插入删除扰动较好，但是在今天看来这个优势也极其微弱了
4）除此之外，红黑树并不比AVL树、SB树、跳表更加优秀
习题1 给定一个数组arr，和两个整数a和b（a<=b）求arr中有多少个子数组，累加和在[a,b]这个范围上返回达标的子数组数量
public static int countRangeSum1(int[] nums, int lower, int upper) {
int n = nums.length;
long[] sums = new long[n + 1];
for (int i = 0; i < n; ++i)
sums[i + 1] = sums[i] + nums[i];
return countWhileMergeSort(sums, 0, n + 1, lower, upper);
}
private static int countWhileMergeSort(long[] sums, int start, int end, int lower, int upper) {
if (end - start <= 1)
return 0;
int mid = (start + end) / 2;
int count = countWhileMergeSort(sums, start, mid, lower, upper)
+ countWhileMergeSort(sums, mid, end, lower, upper);
int j = mid, k = mid, t = mid;
long[] cache = new long[end - start];
for (int i = start, r = 0; i < mid; ++i, ++r) {
while (k < end && sums[k] - sums[i] < lower)
k++;
while (j < end && sums[j] - sums[i] <= upper)
j++;
while (t < end && sums[t] < sums[i])
cache[r++] = sums[t++];
cache[r] = sums[i];
count += j - k;
}
System.arraycopy(cache, 0, sums, start, t - start);
return count;
}
public static class SBTNode {
public long key;
public SBTNode l;
public SBTNode r;
public long size; // 不同key的size
public long all; // 总的size
public SBTNode(long k) {
key = k;
size = 1;
all = 1;
}
}
public static class SizeBalancedTreeSet {
private SBTNode root;
private HashSet<Long> set = new HashSet<>();
private SBTNode rightRotate(SBTNode cur) {
long same = cur.all - (cur.l != null ? cur.l.all : 0) - (cur.r != null ? cur.r.all : 0);
SBTNode leftNode = cur.l;
cur.l = leftNode.r;
leftNode.r = cur;
leftNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
// all modify
leftNode.all = cur.all;
cur.all = (cur.l != null ? cur.l.all : 0) + (cur.r != null ? cur.r.all : 0) + same;
return leftNode;
}
private SBTNode leftRotate(SBTNode cur) {
long same = cur.all - (cur.l != null ? cur.l.all : 0) - (cur.r != null ? cur.r.all : 0);
SBTNode rightNode = cur.r;
cur.r = rightNode.l;
rightNode.l = cur;
rightNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
// all modify
rightNode.all = cur.all;
cur.all = (cur.l != null ? cur.l.all : 0) + (cur.r != null ? cur.r.all : 0) + same;
return rightNode;
}
private SBTNode maintain(SBTNode cur) {
if (cur == null) {
return null;
}
long leftSize = cur.l != null ? cur.l.size : 0;
long leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0;
long leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0;
long rightSize = cur.r != null ? cur.r.size : 0;
long rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0;
long rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0;
if (leftLeftSize > rightSize) {
cur = rightRotate(cur);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (leftRightSize > rightSize) {
cur.l = leftRotate(cur.l);
cur = rightRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (rightRightSize > leftSize) {
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur = maintain(cur);
} else if (rightLeftSize > leftSize) {
cur.r = rightRotate(cur.r);
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
}
return cur;
}
private SBTNode add(SBTNode cur, long key, boolean contains) {
if (cur == null) {
return new SBTNode(key);
} else {
cur.all++;
if (key == cur.key) {
return cur;
} else { // 还在左滑或者右滑
if (!contains) {
cur.size++;
}
if (key < cur.key) {
cur.l = add(cur.l, key, contains);
} else {
cur.r = add(cur.r, key, contains);
}
return maintain(cur);
}
}
}
public void add(long sum) {
boolean contains = set.contains(sum);
root = add(root, sum, contains);
set.add(sum);
}
public long lessKeySize(long key) {
SBTNode cur = root;
long ans = 0;
while (cur != null) {
if (key == cur.key) {
return ans + (cur.l != null ? cur.l.all : 0);
} else if (key < cur.key) {
cur = cur.l;
} else {
ans += cur.all - (cur.r != null ? cur.r.all : 0);
cur = cur.r;
}
}
return ans;
}
// > 7 8...
// <8 ...<=7
public long moreKeySize(long key) {
return root != null ? (root.all - lessKeySize(key + 1)) : 0;
}
}
public static int countRangeSum2(int[] nums, int lower, int upper) {
// 黑盒，加入数字（前缀和），不去重，可以接受重复数字
// < num , 有几个数？
SizeBalancedTreeSet treeSet = new SizeBalancedTreeSet();
long sum = 0;
int ans = 0;
treeSet.add(0);// 一个数都没有的时候，就已经有一个前缀和累加和为0，
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
// [sum - upper, sum - lower]
// [10, 20] ?
// < 10 ? < 21 ?
long a = treeSet.lessKeySize(sum - lower + 1);
long b = treeSet.lessKeySize(sum - upper);
ans += a - b;
treeSet.add(sum);
}
return ans;
}
// for test
public static void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
}
// for test
public static int[] generateArray(int len, int varible) {
int[] arr = new int[len];
for (int i = 0; i < arr.length; i++) {
arr[i] = (int) (Math.random() * varible);
}
return arr;
}
public static void main(String[] args) {
int len = 200;
int varible = 50;
for (int i = 0; i < 10000; i++) {
int[] test = generateArray(len, varible);
int lower = (int) (Math.random() * varible) - (int) (Math.random() * varible);
int upper = lower + (int) (Math.random() * varible);
int ans1 = countRangeSum1(test, lower, upper);
int ans2 = countRangeSum2(test, lower, upper);
if (ans1 != ans2) {
printArray(test);
System.out.println(lower);
System.out.println(upper);
System.out.println(ans1);
System.out.println(ans2);
}
}
}
习题2 有一个滑动窗口（讲过的）：1）L是滑动窗口最左位置、R是滑动窗口最右位置，一开始LR都在数组左侧 2）任何一步都可能R往右动，表示某个数进了窗口 3）任何一步都可能L往右动，表示某个数出了窗口 想知道每一个窗口状态的中位数
public static class SBTNode<K extends Comparable<K>> {
public K key;
public SBTNode<K> l;
public SBTNode<K> r;
public int size;
public SBTNode(K k) {
key = k;
size = 1;
}
}
public static class SizeBalancedTreeMap<K extends Comparable<K>> {
private SBTNode<K> root;
private SBTNode<K> rightRotate(SBTNode<K> cur) {
SBTNode<K> leftNode = cur.l;
cur.l = leftNode.r;
leftNode.r = cur;
leftNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
return leftNode;
}
private SBTNode<K> leftRotate(SBTNode<K> cur) {
SBTNode<K> rightNode = cur.r;
cur.r = rightNode.l;
rightNode.l = cur;
rightNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
return rightNode;
}
private SBTNode<K> maintain(SBTNode<K> cur) {
if (cur == null) {
return null;
}
int leftSize = cur.l != null ? cur.l.size : 0;
int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0;
int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0;
int rightSize = cur.r != null ? cur.r.size : 0;
int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0;
int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0;
if (leftLeftSize > rightSize) {
cur = rightRotate(cur);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (leftRightSize > rightSize) {
cur.l = leftRotate(cur.l);
cur = rightRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (rightRightSize > leftSize) {
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur = maintain(cur);
} else if (rightLeftSize > leftSize) {
cur.r = rightRotate(cur.r);
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
}
return cur;
}
private SBTNode<K> findLastIndex(K key) {
SBTNode<K> pre = root;
SBTNode<K> cur = root;
while (cur != null) {
pre = cur;
if (key.compareTo(cur.key) == 0) {
break;
} else if (key.compareTo(cur.key) < 0) {
cur = cur.l;
} else {
cur = cur.r;
}
}
return pre;
}
private SBTNode<K> add(SBTNode<K> cur, K key) {
if (cur == null) {
return new SBTNode<K>(key);
} else {
cur.size++;
if (key.compareTo(cur.key) < 0) {
cur.l = add(cur.l, key);
} else {
cur.r = add(cur.r, key);
}
return maintain(cur);
}
}
private SBTNode<K> delete(SBTNode<K> cur, K key) {
cur.size--;
if (key.compareTo(cur.key) > 0) {
cur.r = delete(cur.r, key);
} else if (key.compareTo(cur.key) < 0) {
cur.l = delete(cur.l, key);
} else {
if (cur.l == null && cur.r == null) {
// free cur memory -> C++
cur = null;
} else if (cur.l == null && cur.r != null) {
// free cur memory -> C++
cur = cur.r;
} else if (cur.l != null && cur.r == null) {
// free cur memory -> C++
cur = cur.l;
} else {
SBTNode<K> pre = null;
SBTNode<K> des = cur.r;
des.size--;
while (des.l != null) {
pre = des;
des = des.l;
des.size--;
}
if (pre != null) {
pre.l = des.r;
des.r = cur.r;
}
des.l = cur.l;
des.size = des.l.size + (des.r == null ? 0 : des.r.size) + 1;
// free cur memory -> C++
cur = des;
}
}
return cur;
}
private SBTNode<K> getIndex(SBTNode<K> cur, int kth) {
if (kth == (cur.l != null ? cur.l.size : 0) + 1) {
return cur;
} else if (kth <= (cur.l != null ? cur.l.size : 0)) {
return getIndex(cur.l, kth);
} else {
return getIndex(cur.r, kth - (cur.l != null ? cur.l.size : 0) - 1);
}
}
public int size() {
return root == null ? 0 : root.size;
}
public boolean containsKey(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
SBTNode<K> lastNode = findLastIndex(key);
return lastNode != null && key.compareTo(lastNode.key) == 0 ? true : false;
}
public void add(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
SBTNode<K> lastNode = findLastIndex(key);
if (lastNode == null || key.compareTo(lastNode.key) != 0) {
root = add(root, key);
}
}
public void remove(K key) {
if (key == null) {
throw new RuntimeException("invalid parameter.");
}
if (containsKey(key)) {
root = delete(root, key);
}
}
public K getIndexKey(int index) {
if (index < 0 || index >= this.size()) {
throw new RuntimeException("invalid parameter.");
}
return getIndex(root, index + 1).key;
}
}
public static class Node implements Comparable<Node> {
public int index;
public int value;
public Node(int i, int v) {
index = i;
value = v;
}
@Override
public int compareTo(Node o) {
return value != o.value ? Integer.valueOf(value).compareTo(o.value)
: Integer.valueOf(index).compareTo(o.index);
}
}
public static double[] medianSlidingWindow(int[] nums, int k) {
SizeBalancedTreeMap<Node> map = new SizeBalancedTreeMap<>();
for (int i = 0; i < k - 1; i++) {
map.add(new Node(i, nums[i]));
}
double[] ans = new double[nums.length - k + 1];
int index = 0;
for (int i = k - 1; i < nums.length; i++) {
map.add(new Node(i, nums[i]));
if (map.size() % 2 == 0) {
Node upmid = map.getIndexKey(map.size() / 2 - 1);
Node downmid = map.getIndexKey(map.size() / 2);
ans[index++] = ((double) upmid.value + (double) downmid.value) / 2;
} else {
Node mid = map.getIndexKey(map.size() / 2);
ans[index++] = (double) mid.value;
}
map.remove(new Node(i - k + 1, nums[i - k + 1]));
}
return ans;
}
习题3 设计一个结构包含如下两个方法：void add(int index, int num)：把num加入到index位置 int get(int index) ：取出index位置的值 void remove(int index) ：把index位置上的值删除 要求三个方法时间复杂度O(logN)
public static class SBTNode<V> {
public V value;
public SBTNode<V> l;
public SBTNode<V> r;
public int size;
public SBTNode(V v) {
value = v;
size = 1;
}
}
public static class SbtList<V> {
private SBTNode<V> root;
private SBTNode<V> rightRotate(SBTNode<V> cur) {
SBTNode<V> leftNode = cur.l;
cur.l = leftNode.r;
leftNode.r = cur;
leftNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
return leftNode;
}
private SBTNode<V> leftRotate(SBTNode<V> cur) {
SBTNode<V> rightNode = cur.r;
cur.r = rightNode.l;
rightNode.l = cur;
rightNode.size = cur.size;
cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1;
return rightNode;
}
private SBTNode<V> maintain(SBTNode<V> cur) {
if (cur == null) {
return null;
}
int leftSize = cur.l != null ? cur.l.size : 0;
int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0;
int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0;
int rightSize = cur.r != null ? cur.r.size : 0;
int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0;
int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0;
if (leftLeftSize > rightSize) {
cur = rightRotate(cur);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (leftRightSize > rightSize) {
cur.l = leftRotate(cur.l);
cur = rightRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
} else if (rightRightSize > leftSize) {
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur = maintain(cur);
} else if (rightLeftSize > leftSize) {
cur.r = rightRotate(cur.r);
cur = leftRotate(cur);
cur.l = maintain(cur.l);
cur.r = maintain(cur.r);
cur = maintain(cur);
}
return cur;
}
private SBTNode<V> add(SBTNode<V> root, int index, SBTNode<V> cur) {
if (root == null) {
return cur;
}
root.size++;
int leftAndHeadSize = (root.l != null ? root.l.size : 0) + 1;
if (index < leftAndHeadSize) {
root.l = add(root.l, index, cur);
} else {
root.r = add(root.r, index - leftAndHeadSize, cur);
}
root = maintain(root);
return root;
}
private SBTNode<V> remove(SBTNode<V> root, int index) {
root.size--;
int rootIndex = root.l != null ? root.l.size : 0;
if (index != rootIndex) {
if (index < rootIndex) {
root.l = remove(root.l, index);
} else {
root.r = remove(root.r, index - rootIndex - 1);
}
return root;
}
if (root.l == null && root.r == null) {
return null;
}
if (root.l == null) {
return root.r;
}
if (root.r == null) {
return root.l;
}
SBTNode<V> pre = null;
SBTNode<V> suc = root.r;
suc.size--;
while (suc.l != null) {
pre = suc;
suc = suc.l;
suc.size--;
}
if (pre != null) {
pre.l = suc.r;
suc.r = root.r;
}
suc.l = root.l;
suc.size = suc.l.size + (suc.r == null ? 0 : suc.r.size) + 1;
return suc;
}
private SBTNode<V> get(SBTNode<V> root, int index) {
int leftSize = root.l != null ? root.l.size : 0;
if (index < leftSize) {
return get(root.l, index);
} else if (index == leftSize) {
return root;
} else {
return get(root.r, index - leftSize - 1);
}
}
public void add(int index, V num) {
SBTNode<V> cur = new SBTNode<V>(num);
if (root == null) {
root = cur;
} else {
if (index <= root.size) {
root = add(root, index, cur);
}
}
}
public V get(int index) {
SBTNode<V> ans = get(root, index);
return ans.value;
}
public void remove(int index) {
if (index >= 0 && size() > index) {
root = remove(root, index);
}
}
public int size() {
return root == null ? 0 : root.size;
}
}
// 通过以下这个测试，
// 可以很明显的看到LinkedList的插入、删除、get效率不如SbtList
// LinkedList需要找到index所在的位置之后才能插入或者读取，时间复杂度O(N)
// SbtList是平衡搜索二叉树，所以插入或者读取时间复杂度都是O(logN)
public static void main(String[] args) {
// 功能测试
int test = 50000;
int max = 1000000;
boolean pass = true;
ArrayList<Integer> list = new ArrayList<>();
SbtList<Integer> sbtList = new SbtList<>();
for (int i = 0; i < test; i++) {
if (list.size() != sbtList.size()) {
pass = false;
break;
}
if (list.size() > 1 && Math.random() < 0.5) {
int removeIndex = (int) (Math.random() * list.size());
list.remove(removeIndex);
sbtList.remove(removeIndex);
} else {
int randomIndex = (int) (Math.random() * (list.size() + 1));
int randomValue = (int) (Math.random() * (max + 1));
list.add(randomIndex, randomValue);
sbtList.add(randomIndex, randomValue);
}
}
for (int i = 0; i < list.size(); i++) {
if (!list.get(i).equals(sbtList.get(i))) {
pass = false;
break;
}
}
System.out.println("功能测试是否通过 : " + pass);
// 性能测试
test = 500000;
list = new ArrayList<>();
sbtList = new SbtList<>();
long start = 0;
long end = 0;
start = System.currentTimeMillis();
for (int i = 0; i < test; i++) {
int randomIndex = (int) (Math.random() * (list.size() + 1));
int randomValue = (int) (Math.random() * (max + 1));
list.add(randomIndex, randomValue);
}
end = System.currentTimeMillis();
System.out.println("ArrayList插入总时长(毫秒) ： " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < test; i++) {
int randomIndex = (int) (Math.random() * (i + 1));
list.get(randomIndex);
}
end = System.currentTimeMillis();
System.out.println("ArrayList读取总时长(毫秒) : " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < test; i++) {
int randomIndex = (int) (Math.random() * list.size());
list.remove(randomIndex);
}
end = System.currentTimeMillis();
System.out.println("ArrayList删除总时长(毫秒) : " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < test; i++) {
int randomIndex = (int) (Math.random() * (sbtList.size() + 1));
int randomValue = (int) (Math.random() * (max + 1));
sbtList.add(randomIndex, randomValue);
}
end = System.currentTimeMillis();
System.out.println("SbtList插入总时长(毫秒) : " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < test; i++) {
int randomIndex = (int) (Math.random() * (i + 1));
sbtList.get(randomIndex);
}
end = System.currentTimeMillis();
System.out.println("SbtList读取总时长(毫秒) : " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < test; i++) {
int randomIndex = (int) (Math.random() * sbtList.size());
sbtList.remove(randomIndex);
}
end = System.currentTimeMillis();
System.out.println("SbtList删除总时长(毫秒) : " + (end - start));
}
改写有序表的题目核心点
1）分析增加什么数据项可以支持题目
2）有序表一定要保持内部参与排序的key不重复
3）增加这个数据项了，在平衡性调整时，保证这个数据项也能更新正确
4）做到上面3点，剩下就是搜索二叉树怎么实现你想要的接口的问题了