数据库中怎么合并两个有序链表
本篇内容介绍了“数据库中怎么合并两个有序链表”的有关知识，在实际案例的操作过程中，不少人都会遇到这样的困境，接下来就让小编带领大家学习一下如何处理这些情况吧！希望大家仔细阅读，能够学有所成！
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package com.clean.notes;
public class Node {
String data;
Node next;
public Node() {
}
public Node(String data) {
this.data = data;
}
public Node(String data, Node next) {
this.data = data;
this.next = next;
}
public String getData() {
return data;
}
public void setData(String data) {
this.data = data;
}
}package com.clean.notes;
public class MyNode {
//private static Node head = new Node("0", null);
private Node head = null;
// public static void main(String[] args) {
// addNode("1");
// addNode("2");
// insertNode(head,2,"3");
// deleteNode(head,2);
// traverse(head);
// linkListLength(head);
// System.out.println("------------");
// printNodes(head);
// System.out.println("--------------");
// Node prev = reverseLinkList(head);
// printNodes(prev);
// }
public MyNode(String x) {
head = new Node(x,null);
}
/**
* 添加数据作为节点
*
* @param data
*/
public void addNode(String data) {
Node newNode = new Node(data);
Node temp = this.getHead();
while (temp.next != null) {
temp = temp.next;
}
temp.next = newNode;
}
/**
* 从head 遍历链表中的数据
* @param head
*/
public void traverse(Node head) {
Node temp = head;
while (temp != null) {
System.out.println(temp.data);
temp = temp.next;
}
}
/**
* 获取现在链表的长度
* @param head
* @return
*/
public int linkListLength(Node head) {
Node temp = head;
int i = 0;
while (temp != null) {
i ++;
temp = temp.next;
}
int len = i;
System.out.println("length = " + len);
return len;
}
/**
* 添加节点
* @param head
* @param index
* @param data
*/
public void insertNode(Node head, int index, String data) {
int len = linkListLength(head);
if(index < 1 || index > len) {
System.out.println("输入index不合法!");
return;
}
int currentPos = 0;
Node temp = head;
Node newNode = new Node(data);
while (temp != null) {
if((index - 1) == currentPos) {
newNode.next = temp.next;
temp.next = newNode;
return;
}
temp = temp.next;
currentPos ++;
}
}
/**
* 删除节点
* @param head
* @param index
*/
public void deleteNode(Node head, int index) {
int len = linkListLength(head);
if(index < 1 || index > len) {
System.out.println("输入index不合法!");
return;
}
int currentPos = 0;
Node temp = head;
while (temp.next != null) {
if((index - 1) == currentPos) {
Node deteteNode = temp.next;
temp.next = deteteNode.next;
return;
}
temp = temp.next;
currentPos ++;
}
}
public static void printNodes(Node head) {
if(head != null) {
System.out.println(head.data);
printNodes(head.next);
}
}
public Node getHead() {
return this.head;
}
public void setHead(Node head) {
this.head = head;
}
/**
* 从head 遍历链表中的数据
* @param head
*/
public static void traverseString(Node head) {
StringBuffer stringBuffer = new StringBuffer();
Node temp = head;
while (temp != null) {
//System.out.println(temp.data);
stringBuffer.append(temp.data).append("->");
temp = temp.next;
}
String resStr = stringBuffer.toString();
System.out.println(resStr.substring(0,resStr.lastIndexOf("->")));
}
/**
* 实现链表的反转
*
* @param node 链表的头节点
*/
public static Node reverseLinkList(Node node) {
Node prev ;
if (node == null || node.next == null) {
prev = node;
} else {
Node tmp = reverseLinkList(node.next);
node.next.next = node;
node.next = null;
prev = tmp;
}
return prev;
}
}package com.clean.notes;
public class App {
public static void main(String[] args) {
MyNode L1 = new MyNode("x");
MyNode L2 = new MyNode("y");
MyNode L3 = new MyNode("z");
L1.addNode(7 +"");
L1.addNode(10 +"");
L1.addNode(24 +"");
L1.addNode(68 +"");
L2.addNode(9 +"");
L2.addNode(17 +"");
L2.addNode(23 +"");
L2.addNode(50 +"");
L2.addNode(100 +"");
L2.addNode(200 +"");
int sizeA = L1.linkListLength(L1.getHead());
int sizeB = L2.linkListLength(L2.getHead());
System.out.println("sizeA = " + sizeA);
System.out.println("sizeB = " + sizeB);
L1.traverseString(L1.getHead());
L2.traverseString(L2.getHead());
Node temp = L1.getHead();
Node temp2 = L2.getHead();
while (temp.next!= null && temp2.next != null) {
temp = temp.next;
temp2 = temp2.next;
if(temp.data.compareTo(temp2.data) < 0) {
L3.addNode(temp.data);
L3.addNode(temp2.data);
} else {
L3.addNode(temp2.data);
L3.addNode(temp.data);
}
}
while (temp.next != null) {
temp = temp.next;
L3.addNode(temp.data);
}
while (temp2.next != null) {
temp2 = temp2.next;
L3.addNode(temp2.data);
}
int z = L3.linkListLength(L3.getHead());
System.out.println("length L3 = " + z);
L3.traverseString(L3.getHead());
}
}package com.clean.notes;
public class Client {
public static void main(String[] args) {
Client client = new Client();
MyNode L1 = new MyNode("x");
MyNode L2 = new MyNode("y");
L1.addNode(5 +"");
L1.addNode(4 +"");
L1.addNode(8 +"");
L1.addNode(3 +"");
L1.addNode(2 +"");
L1.addNode(1 +"");
L1.addNode(7 +"");
L2.addNode(9 +"");
L2.addNode(1 +"");
L2.addNode(3 +"");
L2.addNode(2 +"");
L2.addNode(1 +"");
L2.addNode(7 +"");
int sizeA = L1.linkListLength(L1.getHead());
int sizeB = L2.linkListLength(L2.getHead());
System.out.println("sizeA = " + sizeA);
System.out.println("sizeB = " + sizeB);
L1.traverseString(L1.getHead());
L2.traverseString(L2.getHead());
int len = client.getLength(L1,L2);
System.out.println("len = " + len);
Node temp = L1.getHead();
System.out.println(temp.next.data);
int j = 0;
while (j < len) {
temp = temp.next;
j ++;
}
Node temp2 = L2.getHead();
while (!temp.data.equalsIgnoreCase(temp2.data)) {
temp = temp.next;
temp2 = temp2.next;
}
System.out.println("the same = " + temp2.data);
}
public int getLength(MyNode h2, MyNode h3) {
int len1 = h2.linkListLength(h2.getHead());
int len2 = h3.linkListLength(h3.getHead());
if(len1 > len2) {
return len1 - len2;
} else {
return len2 - len1;
}
}
}“数据库中怎么合并两个有序链表”的内容就介绍到这里了，感谢大家的阅读。如果想了解更多行业相关的知识可以关注创新互联网站，小编将为大家输出更多高质量的实用文章！
文章题目：数据库中怎么合并两个有序链表
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