Ordered linked list:
Sort by key value. When deleting the link head, the minimum (/maximum) value is deleted. When inserting, the insertion position is searched.
It needs to be compared O(N) when inserting, and the average is O(N/2). When deleting the smallest (/largest) data at the head of the chain, the efficiency is O(1).
If an application requires frequent storage To get (insert/search/delete) the smallest (/largest) data item, then the ordered linked list is a good choice
The priority queue can use the ordered linked list to implement
Insertion sorting of the ordered linked list:
For an unordered array, use an ordered linked list to sort, the comparison time level is still O(N^2)
The copy time level is O(2*N), because the number of copies is small, the first time Put the data into the linked list and move it N times, and then copy it from the linked list to the array, N times again
Every time you insert a new link point, you don’t need to copy and move the data, you only need to change the link domain of one or two link points
import java.util.Arrays; import java.util.Random; /** * 有序链表 对数组进行插入排序 * @author stone */ public class LinkedListInsertSort> { private Link first; //首结点 public LinkedListInsertSort() { } public boolean isEmpty() { return first == null; } public void sortList(T[] ary) { if (ary == null) { return; } //将数组元素插入进链表,以有序链表进行排序 for (T data : ary) { insert(data); } // } public void insert(T data) {// 插入 到 链头, 以从小到大排序 Link newLink = new Link (data); Link current = first, previous = null; while (current != null && data.compareTo(current.data) > 0) { previous = current; current = current.next; } if (previous == null) { first = newLink; } else { previous.next = newLink; } newLink.next = current; } public Link deleteFirst() {//删除 链头 Link temp = first; first = first.next; //变更首结点，为下一结点 return temp; } public Link find(T t) { Link find = first; while (find != null) { if (!find.data.equals(t)) { find = find.next; } else { break; } } return find; } public Link delete(T t) { if (isEmpty()) { return null; } else { if (first.data.equals(t)) { Link temp = first; first = first.next; //变更首结点，为下一结点 return temp; } } Link p = first; Link q = first; while (!p.data.equals(t)) { if (p.next == null) {//表示到链尾还没找到 return null; } else { q = p; p = p.next; } } q.next = p.next; return p; } public void displayList() {//遍历 System.out.println("List (first-->last):"); Link current = first; while (current != null) { current.displayLink(); current = current.next; } } public void displayListReverse() {//反序遍历 Link p = first, q = first.next, t; while (q != null) {//指针反向，遍历的数据顺序向后 t = q.next; //no3 if (p == first) {// 当为原来的头时，头的.next应该置空 p.next = null; } q.next = p;// no3 -> no1 pointer reverse p = q; //start is reverse q = t; //no3 start } //上面循环中的if里，把first.next 置空了, 而当q为null不执行循环时，p就为原来的最且一个数据项，反转后把p赋给first first = p; displayList(); } class Link {//链结点 T data; //数据域 Link next; //后继指针，结点 链域 Link(T data) { this.data = data; } void displayLink() { System.out.println("the data is " + data.toString()); } } public static void main(String[] args) { LinkedListInsertSort list = new LinkedListInsertSort (); Random random = new Random(); int len = 5; Integer[] ary = new Integer[len]; for (int i = 0; i < len; i++) { ary[i] = random.nextInt(1000); } System.out.println("----排序前----"); System.out.println(Arrays.toString(ary)); System.out.println("----链表排序后----"); list.sortList(ary); list.displayList(); } }
----排序前---- [595, 725, 310, 702, 444] ----链表排序后---- List (first-->last): the data is 310 the data is 444 the data is 595 the data is 702 the data is 725
Singly linked list reverse order:
public class SingleLinkedListReverse { public static void main(String[] args) { Node head = new Node(0); Node temp = null; Node cur = null; for (int i = 1; i <= 10; i++) { temp = new Node(i); if (i == 1) { head.setNext(temp); } else { cur.setNext(temp); } cur = temp; }//10.next = null; Node h = head; while (h != null) { System.out.print(h.getData() + "\t"); h = h.getNext(); } System.out.println(); //反转1 // h = Node.reverse1(head); // while (h != null) { // System.out.print(h.getData() + "\t"); // h = h.getNext(); // } //反转2 h = Node.reverse1(head); while (h != null) { System.out.print(h.getData() + "\t"); h = h.getNext(); } } } /* * 单链表的每个节点都含有指向下一个节点属性 */ class Node { Object data;//数据对象 Node next; //下一节点 Node(Object d) { this.data = d; } Node(Object d, Node n) { this.data = d; this.next = n; } public Object getData() { return data; } public void setData(Object data) { this.data = data; } public Node getNext() { return next; } public void setNext(Node next) { this.next = next; } //方法1 head被重置 static Node reverse1(Node head) { Node p = null; //反转后新的 头 Node q = head; //轮换结果：012,123,234,.... 10 null null while (head.next != null) { p = head.next; // 第1个 换成第2个 这时p表示原始序列头中的next head.next = p.next; // 第2个 换成第3个 p.next = q; //已经跑到第1位置的原第2个的下一个 就要变成 原第1个 q = p; //新的第1个 要变成 当前第一个 } return p; } //方法2 head没重置 static Node reverse2(Node head) { //将中间节点的指针指向前一个节点之后仍然可以继续向后遍历链表 Node p1 = head, p2 = head.next, p3; // 前 中 后 //轮换结果 ：012, 123, 234, 345, 456.... 9 10 null while (p2 != null) { p3 = p2.next; p2.next = p1; //指向后 变 指向前 p1 = p2; //2、3向前挪 p2 = p3; } head.next = null;//head没变，当输出到0时，再请求0.next 为1 return p1; } }
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