数论(树形结构、二叉树、二叉搜索树、红黑树、Btree、B+Tree、赫夫曼树、堆树)
树形结构概念
在树形结构里面重要的术语:
-
结点:树里面的元素。
-
父子关系:结点之间相连的边
-
子树:当结点大于1时,其余的结点分为的互不相交的集合称为子树
-
度:一个结点拥有的子树数量称为结点的度
-
叶子:度为0的结点
-
孩子:结点的子树的根称为孩子结点
-
双亲:和孩子结点对应
-
兄弟:同一个双亲结点
-
森林:由N个互不相交的树构成深林
-
结点的高度:结点到叶子结点的最长路径
-
结点的深度:根结点到该结点的边个数
-
结点的层数:结点的深度加1
-
树的高度:根结点的高度
二叉树
Binary Tree: 一种特殊的树形结构,每个节点至多只有两颗子树
在二叉树的第N层上至多有2^(N-1)
个结点。最多有2^N-1
个结点个数。
分类:
- 满二叉树:除叶子结点外,每个结点都有左右两个子结点。
- 完全二叉树:除最后一层外,其他的结点个数必须达到最大,并且最后一层结点都连续靠左排列。
思考
为什么要分满二叉树和完全二叉树呢?因为通过定义可以看出,完全二叉树只是满二叉树里面的一个子集
数组:性能高效,如果不是完全二叉树浪费空间
链表:也可以实现,性能没有数组高
二叉树遍历
- 重要口诀:根节点输出!子树
- 前序:根 左 右 (A B C D E F G H K)
- 中序:左 根 右 (B C D A E F G H K)
- 后序:左 右 根 (B C D E F G H K A)
/*** 二叉树--前中后链式遍历* 前: A B C D E F G H K* 中: B C D A E F G H K* 后: B C D E F G H K A* 层:* A* B E* C F* D G* H K*/
@Data
class TreeNode {private char data;private TreeNode left;private TreeNode right;public TreeNode(char data, TreeNode left, TreeNode right) {this.data = data;this.left = left;this.right = right;}
}public class BinaryTree {public void print(TreeNode node) {System.out.print(node.getData() + " ");}public void pre(TreeNode root) { //前序:根(输出) 左 右 A B C D E F G H Kprint(root);if (root.getLeft() != null) {pre(root.getLeft()); //认为是子树,分解子问题}if (root.getRight() != null) {pre(root.getRight());}}public void in(TreeNode root) { //中序:左 根(输出) 右 B C D A E F G H Kif (root.getLeft() != null) {pre(root.getLeft()); //认为是子树,分解子问题}print(root);if (root.getRight() != null) {pre(root.getRight());}}public void post(TreeNode root) { //后序:左 右 根(输出) B C D E F G H K Aif (root.getLeft() != null) {pre(root.getLeft()); //认为是子树,分解子问题}if (root.getRight() != null) {pre(root.getRight());}print(root);}public List<List<Character>> level(TreeNode root) { //层次遍历if (root == null) return Collections.EMPTY_LIST;List<List<Character>> res = new ArrayList<>();Queue<TreeNode> queue = new LinkedList<>();queue.add(root);while (!queue.isEmpty()) {List<Character> raw = new ArrayList<>();int size = queue.size();for (int i = 0; i < size; i++) {TreeNode item = queue.poll();raw.add(item.getData());if (item.getLeft() != null) {queue.add(item.getLeft());}if (item.getRight() != null) {queue.add(item.getRight());}}res.add(raw);}return res;}public static void main(String[] args) {TreeNode D = new TreeNode('D', null,null);TreeNode H = new TreeNode('H', null,null);TreeNode K = new TreeNode('K', null,null);TreeNode C = new TreeNode('C', D,null);TreeNode G = new TreeNode('G', H,K);TreeNode B = new TreeNode('B', null,C);TreeNode F = new TreeNode('F', G,null);TreeNode E = new TreeNode('E', F,null);TreeNode A = new TreeNode('A', B,E);BinaryTree tree = new BinaryTree();System.out.print("前: ");tree.pre(A);System.out.println();System.out.print("中: ");tree.in(A);System.out.println();System.out.print("后: ");tree.post(A);System.out.println();System.out.println("层: ");List<List<Character>> res = tree.level(A);for (List<Character> re : res) {for (Character character : re) {System.out.print(character + " ");}System.out.println();}}
}
二叉搜索树(二叉查找树、二叉排序数)
- 如果它的左子树不为空,则左子树上结点的值都小于根结点
- 如果它的右子树不为空,则右子树上结点的值都大于根结点
- 子树同样也要遵循以上两点
中序遍历 ---- 左 根(输出) 右:0 3 4 5 6 8
性能分析
- 查找logn
- 插入nlogn
- 删除
-
- 要删除的结点是叶子结点 O(1)
-
- 要删除的结点只有一个子树(左或者右)O(1)
-
- 要删除的结点有两颗子树:找后继结点,而且后继结点的左子树一定为空 logn
-
/*** 二叉搜索树 增删改查*/
class BinaryNodeTeacher {int data;BinaryNodeTeacher left;BinaryNodeTeacher right;BinaryNodeTeacher parent;public BinaryNodeTeacher(int data) {this.data = data;this.left = null;this.right = null;this.parent = null;}
}public class BinarySearchTreeTeacher {public BinaryNodeTeacher find(BinaryNodeTeacher root, int key) {BinaryNodeTeacher current = root;while (current != null) {if (key < current.data) {current = current.left;} else if (key > current.data) {current = current.right;} else {return current;}}return null;}public void insert(BinaryNodeTeacher root, int data) {if (root.data < data) {if (root.right != null) {insert(root.right, data);} else {BinaryNodeTeacher newNode = new BinaryNodeTeacher(data);newNode.parent = root;root.right = newNode;}} else {if (root.left != null) {insert(root.left, data);} else {BinaryNodeTeacher newNode = new BinaryNodeTeacher(data);newNode.parent = root;root.left = newNode;}}}public BinaryNodeTeacher finSuccessor(BinaryNodeTeacher node) { // 查找node的后继节点if (node.right == null) { // 表示没有右边 那就没有后继return node;}BinaryNodeTeacher cur = node.right;BinaryNodeTeacher pre = node.right; // 开一个额外的空间 用来返回后继节点,因为我们要找到为空的时候,那么其实返回的是上一个节点while (cur != null) {pre = cur;cur = cur.left; // 注意后继节点是要往左边找,因为右边的肯定比左边的大,我们要找的是第一个比根节点小的,所以只能往左边}return pre; // 因为cur会变成null,实际我们是要cur的上一个点,所以就是pre来代替}public BinaryNodeTeacher remove(BinaryNodeTeacher root, int data) { // 删除dataBinaryNodeTeacher delNode = find(root, data);if (delNode == null) {System.out.println("要删除的值不在树中");return root;}// 1.删除的点没有左右子树if (delNode.left == null && delNode.right == null) {if (delNode == root) {root = null;} else if (delNode.parent.data < delNode.data) { // 说明删除的点是右子节点delNode.parent.right = null;} else {delNode.parent.left = null;}} else if (delNode.left != null && delNode.right != null) { // 2.删除的节点有两颗子节点BinaryNodeTeacher successor = finSuccessor(delNode); // 先找的后继节点// 后继节点和删除节点进行交换,首先后继节点的左节点是肯定为空的successor.left = delNode.left; // 后继的左边变为删除的左边successor.left.parent = successor; // 删除点的左边parent指向后继节点// 再来看后继节点的右边if (successor.right != null && successor.parent != delNode) { // 后继节点有右边,这其实就是下面情况3的第一种successor.right.parent = successor.parent;successor.parent.left = successor.right;successor.right = delNode.right;successor.right.parent = successor;}else if(successor.right == null) { //如果后继节点没有右边,那其实就是情况1,没有左右子树if(successor.parent != delNode) { //如果后继节点的parent不等于删除的点 那么就需要把删除的右子树赋值给后继节点successor.parent.left = null; //注意原来的后继节点上的引用要删掉,否则会死循环successor.right = delNode.right;successor.right.parent = successor;}}// 替换做完接下来就要删除节点了if (delNode == root) {successor.parent = null;root = successor;return root;}successor.parent = delNode.parent;if (delNode.data > delNode.parent.data) { // 删除的点在右边,关联右子树delNode.parent.right = successor;} else {delNode.parent.left = successor;}} else { // 3.删除点有一个节点if (delNode.right != null) { // 有右节点if (delNode == root) {root = delNode.right;return root;}delNode.right.parent = delNode.parent; // 把右节点的parent指向删除点的parent// 关联父节点的左右子树if (delNode.data < delNode.parent.data) { // 删除的点在左边delNode.parent.left = delNode.right;} else {delNode.parent.right = delNode.right;}} else {if (delNode == root) {root = delNode.left;return root;}delNode.left.parent = delNode.parent;if (delNode.data < delNode.parent.data) {delNode.parent.left = delNode.left;} else {delNode.parent.right = delNode.left;}}}return root;}public void inOrde(BinaryNodeTeacher root) {if (root != null) {inOrde(root.left);System.out.print(root.data);inOrde(root.right);}}// 用于获得树的层数public int getTreeDepth(BinaryNodeTeacher root) {return root == null ? 0 : (1 + Math.max(getTreeDepth(root.left), getTreeDepth(root.right)));}/**** 测试用例* 15* 10* 19* 8* 13* 16* 28* 5* 9* 12* 14* 20* 30* -1* 删除:15 8 5 10 12 19 16 14 30 9 13 20 28** 15* / \* 10 19* / \ / \* 8 13 16 28* / \ / \ / \* 5 9 12 14 20 30*/public static void main(String[] args) {BinarySearchTreeTeacher binarySearchTree = new BinarySearchTreeTeacher();BinaryNodeTeacher root = null;Scanner cin = new Scanner(System.in);int t = 1;System.out.println("二叉搜索树假定不存重复的子节点,重复可用链表处理,请注意~~");System.out.println("请输入根节点:");int rootData = cin.nextInt();root = new BinaryNodeTeacher(rootData);System.out.println("请输入第" + t + "个点:输入-1表示结束");while (true) { //int data = cin.nextInt();if (data == -1)break;binarySearchTree.insert(root, data);t++;System.out.println("请输入第" + t + "个点:输入-1表示结束");}binarySearchTree.show(root); //找的别人写的打印二叉树形结构,感觉还不错,可以更加清晰System.out.println("删除测试:");while(true) {System.out.println("请输入要删除的点:-1表示结束");int key = cin.nextInt();root = binarySearchTree.remove(root, key);binarySearchTree.show(root);if(root == null) {System.out.println("树已经没有数据了~~");break;}}}private void writeArray(BinaryNodeTeacher currNode, int rowIndex, int columnIndex, String[][] res, int treeDepth) {// 保证输入的树不为空if (currNode == null)return;// 先将当前节点保存到二维数组中res[rowIndex][columnIndex] = String.valueOf(currNode.data);// 计算当前位于树的第几层int currLevel = ((rowIndex + 1) / 2);// 若到了最后一层,则返回if (currLevel == treeDepth)return;// 计算当前行到下一行,每个元素之间的间隔(下一行的列索引与当前元素的列索引之间的间隔)int gap = treeDepth - currLevel - 1;// 对左儿子进行判断,若有左儿子,则记录相应的"/"与左儿子的值if (currNode.left != null) {res[rowIndex + 1][columnIndex - gap] = "/";writeArray(currNode.left, rowIndex + 2, columnIndex - gap * 2, res, treeDepth);}// 对右儿子进行判断,若有右儿子,则记录相应的"\"与右儿子的值if (currNode.right != null) {res[rowIndex + 1][columnIndex + gap] = "\\";writeArray(currNode.right, rowIndex + 2, columnIndex + gap * 2, res, treeDepth);}}public void show(BinaryNodeTeacher root) {if (root == null) {System.out.println("EMPTY!");return ;}// 得到树的深度int treeDepth = getTreeDepth(root);// 最后一行的宽度为2的(n - 1)次方乘3,再加1// 作为整个二维数组的宽度int arrayHeight = treeDepth * 2 - 1;int arrayWidth = (2 << (treeDepth - 2)) * 3 + 1;// 用一个字符串数组来存储每个位置应显示的元素String[][] res = new String[arrayHeight][arrayWidth];// 对数组进行初始化,默认为一个空格for (int i = 0; i < arrayHeight; i++) {for (int j = 0; j < arrayWidth; j++) {res[i][j] = " ";}}// 从根节点开始,递归处理整个树writeArray(root, 0, arrayWidth / 2, res, treeDepth);// 此时,已经将所有需要显示的元素储存到了二维数组中,将其拼接并打印即可for (String[] line : res) {StringBuilder sb = new StringBuilder();for (int i = 0; i < line.length; i++) {sb.append(line[i]);if (line[i].length() > 1 && i <= line.length - 1) {i += line[i].length() > 4 ? 2 : line[i].length() - 1;}}System.out.println(sb.toString());}}}
红黑树(实验室:AVL平衡二叉树)二叉搜索树退化成链表
红黑树的性质:
- 每个结点不是红色就是黑色
- 不可能有连在一起的红色结点(黑色的就可以),每个叶子节点都是黑色的空节点(NIL),也就是说,叶子节点不存储数据
- 根结点都是黑色 root
- 每个节点,从该节点到达其可达叶子节点的所有路径,都包含相同数目的黑色节点
插入的时候旋转和颜色变换规则:
- 变颜色的情况:当前结点的父亲是红色,且它的祖父结点的另一个子结点
也是红色。(叔叔结点):
(1)把父节点设为黑色
(2)把叔叔也设为黑色
(3)把祖父也就是父亲的父亲设为红色(爷爷)
(4)把指针定义到祖父结点(爷爷)设为当前要操作的. - 左旋:当前父结点是红色,叔叔是黑色的时候,且当前的结点是右子树。左旋
以父结点作为左旋。指针变换到父亲结点 - 右旋:当前父结点是红色,叔叔是黑色的时候,且当前的结点是左子树。右旋
(1)把父结点变为黑色
(2)把祖父结点变为红色 (爷爷)
(3)以祖父结点旋转(爷爷)
左旋
右旋
红黑树的应用
- HashMap
- TreeMap
- Windows底层:查找
- Linux进程调度,nginx等
COPY
public class RedBlackTree {private final int R = 0;private final int B = 1;private class Node {int key = -1;int color = B; // 颜色Node left = nil; // nil表示的是叶子结点Node right = nil;Node p = nil;Node(int key) {this.key = key;}@Overridepublic String toString() {return "Node [key=" + key + ", color=" + color + ", left=" + left.key + ", right=" + right.key + ", p=" + p.key + "]" + "\r\n";}}private final Node nil = new Node(-1);private Node root = nil;public void printTree(Node node) {if (node == nil) {return;}printTree(node.left);System.out.print(node.toString());printTree(node.right);}private void insert(Node node) {Node temp = root;if (root == nil) {root = node;node.color = B;node.p = nil;} else {node.color = R;while (true) {if (node.key < temp.key) {if (temp.left == nil) {temp.left = node;node.p = temp;break;} else {temp = temp.left;}} else if (node.key >= temp.key) {if (temp.right == nil) {temp.right = node;node.p = temp;break;} else {temp = temp.right;}}}fixTree(node);}}private void fixTree(Node node) {while (node.p.color == R) {Node y = nil;if (node.p == node.p.p.left) {y = node.p.p.right;if (y != nil && y.color == R) {node.p.color = B;y.color = B;node.p.p.color = R;node = node.p.p;continue;}if (node == node.p.right) {node = node.p;rotateLeft(node);}node.p.color = B;node.p.p.color = R;rotateRight(node.p.p);} else {y = node.p.p.left;if (y != nil && y.color == R) {node.p.color = B;y.color = B;node.p.p.color = R;node = node.p.p;continue;}if (node == node.p.left) {node = node.p;rotateRight(node);}node.p.color = B;node.p.p.color = R;rotateLeft(node.p.p);}}root.color = B;}void rotateLeft(Node node) {if (node.p != nil) {if (node == node.p.left) {node.p.left = node.right;} else {node.p.right = node.right;}node.right.p = node.p;node.p = node.right;if (node.right.left != nil) {node.right.left.p = node;}node.right = node.right.left;node.p.left = node;} else {Node right = root.right;root.right = right.left;right.left.p = root;root.p = right;right.left = root;right.p = nil;root = right;}}void rotateRight(Node node) {if (node.p != nil) {if (node == node.p.left) {node.p.left = node.left;} else {node.p.right = node.left;}node.left.p = node.p;node.p = node.left;if (node.left.right != nil) {node.left.right.p = node;}node.left = node.left.right;node.p.right = node;} else {Node left = root.left;root.left = root.left.right;left.right.p = root;root.p = left;left.right = root;left.p = nil;root = left;}}public void creatTree() {int data[]= {23,32,15,221,3};Node node;System.out.println(Arrays.toString(data));for(int i = 0 ; i < data.length ; i++) {node = new Node(data[i]);insert(node);}printTree(root);}/*** [23, 32, 15, 221, 3]* Node [key=3, color=0, left=-1, right=-1, p=15]* Node [key=15, color=1, left=3, right=-1, p=23]* Node [key=23, color=1, left=15, right=32, p=-1]* Node [key=32, color=1, left=-1, right=221, p=23]* Node [key=221, color=0, left=-1, right=-1, p=32]*/public static void main(String[] args) {RedBlackTree bst = new RedBlackTree();bst.creatTree();}
}
Btree&B+Tree
B-Tree和B+Tree的区别
- B-tree所有的节点都会存数据
- b-tree叶子节点没有链表
数据库索引是什么样的数据结构呢?它为什么又能这么高效的查找呢?究竟使用了什么样的算法呢?
select * from table where id = 10
select * from table where id > 12
select * from table where id > 12 and id < 20
改造二叉搜索树:
能解决我们上面所有的sql语句;
效率 logn
2^32=21亿;
IO:指的是从磁盘读取数据。32层就要读取32次。CPU,内存,IO;
IO从磁盘读一次会读多少数据?计算机组成原理。Page的。页,4KB
Int占多少空间?4B
思考
问题1:搜索效率:32次 (B+Tree 多叉树)
问题2:查询次数: (B+Tree 范围 )
问题3:磁盘IO:解决这个问题;(B+Tree 只有叶子节点存储数据地址)
B+Tree 数据结构
Mysql 如何利用B+Tree 解决问题
Mysql 通过页大小决定,一般是16kb,一个bigint主键类型创建索引消耗的空间是多少?
int 8 字节,指针一个算4字节,一页的节点:16kb/(8+8)=1k 键值+指针
三阶:102410241024=10 7374 1824
如何正确的建立索引:
- 索引不能太多,因为B+tree的插入和删除是要维护的,太多的索引会导致插入变慢。
- 建了索引的字段不能使用like ‘%%’否则是失效的
- 建索引的字段类型不能太大,字段越小阶数就越大,效率就越高,int 和 bigint,varchar(10),varchar(100),text,lontext;B+Tree。全文索引
- 建索引的字段值不能太多一样的,数学里面有个叫什么散列多一些(离散),比如我们把性别建索引会出现啥情况?左边都是一样的值 过滤不了一半。User sex单独建索引 0 1
- 联合索引的最左匹配原则。Select * from user where name = ‘mx’ and id = 1 我的对( id,name)建的索引,mysql解析的时候会自动优化。
Select * from user where name = ‘mx’ and age=10
我的对( id,name,age)建的索引 - NOT IN 是不会走索引的 not in (1,2,3) In的值太多 mysql会报错的
赫夫曼树(哈夫曼树、哈夫曼编码、前缀编码)-- 压缩软件、通信电报
电报的设计:
1.电报加密后越短越好,发送快
2.破解难
3.解码容易
4.换加密树也要快
5.可逆的
计算下面三颗二叉树的带权路径长度总和:
WPL(a):7*2+5*2+2*2+4*2=36()
WPL(b):7*3+5*3+2*1+4*2=46()
WPL(c):7*1+5*2+2*3+4*3=35()
左节点的边设置为0
右节点的边设置为1
© 哈夫曼编码就是
A:0
B:10
C:110
D:111
构建哈夫曼树:
1.每次取数值最小的两个节点,将之组成为一颗子树。
2.移除原来的两个点
3.然后将组成的子树放入原来的序列中
4.重复执行1 2 3 直到只剩最后一个点
/*** 赫夫曼树*/
class HuffmanNode implements Comparable<HuffmanNode> {String chars;int fre; //频率 权重HuffmanNode parent;HuffmanNode left;HuffmanNode right;@Overridepublic int compareTo(HuffmanNode o) {return this.fre - o.fre;}
}public class HuffmanTree {HuffmanNode root;List<HuffmanNode> leafs; //叶子节点Map<Character, Integer> weights; //叶子节点Map<Character, String> charmap;Map<String, Character> mapchar;public HuffmanTree(Map<Character,Integer> weights) {this.weights = weights;leafs = new ArrayList<>();charmap = new HashMap<>();mapchar = new HashMap<>();}public void code() {for (HuffmanNode node : leafs) {Character c = new Character(node.chars.charAt(0));HuffmanNode current = node;String code = "";do {if (current.parent != null && current == current.parent.left) { //leftcode = "0" + code;} else {code = "1" + code;}current = current.parent;} while (current.parent != null);charmap.put(c, code);mapchar.put(code, c);}}public void createTree() {Character keys[] = weights.keySet().toArray(new Character[0]);PriorityQueue<HuffmanNode> priorityQueue = new PriorityQueue<>();for (Character key : keys) {HuffmanNode huffmanNode = new HuffmanNode();huffmanNode.chars = key.toString();huffmanNode.fre = weights.get(key);priorityQueue.add(huffmanNode);leafs.add(huffmanNode);}int len = priorityQueue.size();for (int i = 1; i <= len-1; i++) {HuffmanNode n1 = priorityQueue.poll();HuffmanNode n2 = priorityQueue.poll();HuffmanNode newNode = new HuffmanNode();newNode.fre = n1.fre + n2.fre;newNode.chars = n1.chars + n2.chars;newNode.left = n1;newNode.right = n2;n1.parent = newNode;n2.parent = newNode;priorityQueue.add(newNode);}root = priorityQueue.poll();}public String encode(String body) {StringBuilder builder = new StringBuilder();for (char c : body.toCharArray()) {builder.append(charmap.get(c));}return builder.toString();}public String decode(String body) {StringBuilder builder = new StringBuilder();while (!body.equals("")) {for (String code : mapchar.keySet()) {if (body.startsWith(code)) {body = body.replaceFirst(code,"");builder.append(mapchar.get(code));}}}return builder.toString();}/*** a : 10110* b : 01* c : 1010* d : 00* e : 11* f : 10111* g : 100* encode: 0010111101111010* decode: dffc*/public static void main(String[] args) {Map<Character, Integer> weights = new HashMap<>();weights.put('a',3);weights.put('b',24);weights.put('c',6);weights.put('d',20);weights.put('e',34);weights.put('f',4);weights.put('g',12);HuffmanTree huffmanTree = new HuffmanTree(weights);huffmanTree.createTree();huffmanTree.code();for (Map.Entry<Character, String> entry : huffmanTree.charmap.entrySet()) {System.out.println(entry.getKey() + " : " + entry.getValue());}String encode = huffmanTree.encode("dffc");System.out.println("encode: " + encode);String decode = huffmanTree.decode(encode);System.out.println("decode: " + decode);}
}
堆树
堆的插入有两种实现方式:
- 从下往上
- 从上往下
其插入过程就叫做堆化
/*** 堆树** 建堆* 排序** 1. 优先级队列问题: 删除最大的* 2. top n 热搜排行榜问题:1000万的数字* 3. 定时器 堆顶* 4. 给你1亿不重复的数字,求出top10,前10最大的数字,还可动态添加*/
public class HeapTree {//建大顶堆public static void maxHeap(int data[], int start, int end) {int parent = start;int son = parent * 2 + 1; //下标从0开始+1while (son < end) {int temp = son;//temp表示的是 我们左右节点大的那一个if (son + 1 < end && data[son] < data[son + 1]) {temp = son + 1;}//比较左右节点大的那一个temp和父节点比较大小if (data[parent] > data[temp]) {return;} else {int t = data[parent];data[parent] = data[temp];data[temp] = t;parent = temp; //继续堆化son = parent * 2 + 1;}}}public static void heapSort(int data[]) {int len = data.length;//从后向上建//建堆从哪里开始 最后一个的父元素开始(len/2 - 1)for (int i = len/2 - 1; i >= 0; i--) { //nlog(n)maxHeap(data, i, len);}//从上向下建//最后一个数和第一个数交换for (int i = len - 1; i > 0; i--) { //nlog(n)int temp = data[0];data[0] = data[i];data[i] = temp;maxHeap(data, 0, i);}}/*** Arrays: [1, 3, 4, 7, 8, 14, 17, 20, 25]*/public static void main(String[] args) {int data[] = {8, 4, 20, 7, 3, 1, 25, 14, 17};heapSort(data);System.out.printf("Arrays: " + Arrays.toString(data));}}