链表问题 反转单双链表

要求：

链表长度N，时间复杂度为O(N)，额外空间复杂度为O(1)。

思路总结：

用了四个辅助节点，pre, end, cur, next，代码比较繁杂，但是易于理解。  

收获：

单链表->null。

null<-/->双链表->null。

//反转单双链表
package com.company;

public class Main {
    public static class Node {
        public int value;
        public Node next;
        public Node() { this.next = null; }; //初始化一个空链表
        public Node(int value) { this.value = value; }
    }

    public static class DuLNode {
        public int value;
        public DuLNode next;
        public DuLNode prior;
        public DuLNode() { this.next = null; this.prior = null; }; //初始化一个空链表
        public DuLNode(int value) { this.value = value; }
    }

    public static Node CreatNewLinkedList(Node head, int[] a) {
        //System.out.print("List is: ");
        Node r = head;
        for (int i : a) {
            Node p = new Node(i);
            r.next = p;
            p.next = null;
            r = r.next;
            //System.out.print(p.value + " ");
        }
        //r.next = null;
        //System.out.println();
        return head.next;
    }//尾插法

    public static DuLNode CreatNewDuLLinkedList(DuLNode head, int[] a) {
        DuLNode r = head;
        for (int i : a) {
            DuLNode p = new DuLNode(i);
            r.next = p;
            //p.next = null;
            p.prior = r;
            r = r.next;
        }
        //r.next = null;
        head.next.prior = null;
        return head.next;
    }//尾插法

    public static Node ReverseLinkedList(Node node) {
        Node pre = node;
        Node cur = node.next;
        Node next = node.next.next;
        Node end = node;
        while (next != null) {
            end.next = next;
            cur.next = pre;
            pre = cur;
            cur = next;
            next = next.next;
            if (next == null) {
                end.next = null;
                cur.next = pre;
                pre = cur;
            }
            //System.out.println(1);
            //node = null;
        }
        return pre;
    }

    public static DuLNode ReverseDuLLinkedList(DuLNode node) {
        DuLNode pre = node;
        DuLNode cur = node.next;
        DuLNode next = node.next.next;
        DuLNode end = node;
        while (next != null) {
            end.next = next;
            next.prior = end;
            cur.next = pre;
            pre.prior = cur;
            pre = cur;
            cur = next;
            next = next.next;
            if (next == null) {
                end.next = null;
                cur.next = pre;
                pre.prior = cur;
                cur.prior = null;
                pre = cur;
            }
            //System.out.println(1);
            //node = null;
        }
        return pre;

    }

    public static void PrintLinkedList(Node node) {
        System.out.print("List is: ");
        while (node != null) {
            System.out.print(node.value + " ");
            node = node.next;
        }
        System.out.println();
    }

    public static void PrintLinkedList(DuLNode node) {
        System.out.print("DuLList is: ");
        DuLNode end = new DuLNode();
        while (node != null) {
            System.out.print(node.value + " ");
            end = node;
            node = node.next;
        }
        System.out.print("| ");
        while (end != null) {
            System.out.print(end.value + " ");
            end = end.prior;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        int[] a = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
        Node head = new Node();
        DuLNode duLNode = new DuLNode();

        head = CreatNewLinkedList(head, a);
        PrintLinkedList(head);
        head = ReverseLinkedList(head);
        PrintLinkedList(head);
        System.out.println("===================");

        duLNode = CreatNewDuLLinkedList(duLNode, a);
        //int aaa = 2;
        //while (aaa-- != 0) {
            PrintLinkedList(duLNode);
        //}
        duLNode = ReverseDuLLinkedList(duLNode);
        PrintLinkedList(duLNode);
    }
}