反转单链表和双链表

Problem:
　　反转单向和双向链表
　　【题目】 分别实现反转单向链表和反转双向链表的函数。
　　【要求】 如果链表长度为N，时间复杂度要求为O(N)，额外空间
　　复杂度要求为O(1)

Solution:
　　学会使用指针

Code:

　　

#pragma once

#include <iostream>

using namespace std;

struct Node
{
    int val;
    Node* next;
    Node(int a = 0) :val(a), next(NULL) {}
};

void ReverSingleList(Node*& head)//反转单向链表
{
    if (head->next == NULL || head->next->next == NULL)
        return;
    Node *p1, *p2, *p3;
    p1 = head->next;
    p2 = p1->next;
    p3 = p2->next;
    p1->next = NULL;
    while (p3)
    {
        p2->next = p1;
        p1 = p2;
        p2 = p3;
        p3 = p3->next;
    }
    p2->next = p1;
    head->next = p2;
}


struct DNode
{
    int val;
    DNode* next;
    DNode* pre;
    DNode(int a = 0) :val(a), next(NULL),pre(NULL) {}
};

void ReverDoubleList(DNode*& head)//反转双向链表，方法与单向链表一样
{
    if (head->next == NULL || head->next->next == NULL)
        return;
    DNode *p1, *p2, *p3;
    p1 = head->next;
    p2 = p1->next;
    p3 = p2->next;
    p1->next = NULL;
    p1->pre = p2;
    while (p3)
    {
        p2->next = p1;
        p1->pre = p2;
        p2->pre = p3;
        p1 = p2;
        p2 = p3;
        p3 = p3->next;
    }
    p2->next = p1;
    p2->pre = head;
    head->next = p2;
}

void Test()
{
    int a[6] = { 1,2,3,4,5,6 };
    Node* head = new Node(-1);
    Node* p = head;
    for (auto n : a)
    {
        Node* q = new Node(n);
        p->next = q;
        p = q;
    }
    p->next = NULL;
    
    p = head->next;
    cout << "原链表为： ";
    while (p)
    {
        cout << p->val << "->";
        p = p->next;
    }

    ReverSingleList(head);
    p = head->next;
    cout << endl << "*******************" << endl << "反转后的链表为： ";
    while (p)
    {
        cout << p->val << "->";
        p = p->next;
    }
    cout << endl << "=============================" << endl;
    cout << "=============================" << endl;
    cout << "=============================" << endl;


    int b[6] = { 1,2,3,4,5,6 };
    DNode* head2 = new DNode(-1);
    DNode* p2 = head2;
    for (auto n : b)
    {
        DNode* q2 = new DNode(n);
        p2->next = q2;
        q2->pre = p2;
        p2 = q2;
    }
    p2->next = NULL;

    p2 = head2->next;
    cout << "原链表为的顺为： ";
    while (p2->next)
    {
        cout << p2->val << "->";
        p2 = p2->next;
    }
    cout << p2->val << "->";
    cout << endl << "*******************" << endl << "原链表为的逆为： ";
    while (p2->pre)
    {
        cout << p2->val << "->";
        p2 = p2->pre;
    }

    ReverDoubleList(head2);
    p2 = head2->next;
    cout << endl << "*******************" << endl << "反转后的双向链表的顺为： ";
    while (p2->next)
    {
        cout << p2->val << "->";
        p2 = p2->next;
    }
    cout << p2->val << "->";
    cout << endl << "*******************" << endl << "反转后的双向链表的逆为： ";
    while (p2->pre)
    {
        cout << p2->val << "->";
        p2 = p2->pre;
    }
    cout << endl << "=============================" << endl;
}