反转单向和双向链表

问题描述：

反转单向链表和双向链表

要求：

如果链表长度为N，时间复杂度要求为O(N),额外的空间复杂度要求为O(1)

图解算法

代码实现

#include <iostream>

using namespace std;

struct SingleNode
{
    int value;
    SingleNode *next;
};

struct DoubleNode
{
    int value;
    DoubleNode *pre;
    DoubleNode *next;
};

SingleNode* reverseList(SingleNode *head)
{
    SingleNode *pre = NULL;
    SingleNode *next = NULL;
    //head为当前，pre为之前，next为之后
    while(NULL != head)
    {
        next = head->next;
        head->next = pre;
        pre = head;
        head = next;
    }
    return pre;
}

DoubleNode* reverseList(DoubleNode *head)
{
    DoubleNode *pre = NULL;
    DoubleNode *next = NULL;

    while(NULL != head)
    {
        next = head->next;
        head->next = pre;
        head->pre = next;    //多了一步前置处理
        pre = head;
        head = next;
    }
    return pre;
}

void printList(SingleNode *head)
{
    while(NULL != head)
    {
        cout << head->value << " ";
        head = head->next;
    }
    cout << endl;
}

void printList(DoubleNode *head)
{
    while(NULL != head)
    {
        cout << head->value << " ";
        head = head->next;
    }
    cout << endl;
}

int main()
{
    SingleNode *shead = NULL;
    SingleNode *sptr;
    DoubleNode *dhead = NULL;
    DoubleNode *dptr;
    for(int i =0;i<10;i++)
    {
        if(NULL == shead && NULL == dhead)
        {    
            shead = new SingleNode;
            shead->value = i;
            shead->next = NULL;
            sptr = shead;
            
            dhead = new DoubleNode;
            dhead->value = i;
            dhead->next = NULL;
            dhead->pre = NULL;
            dptr = dhead;
            continue;
        }
        sptr->next = new SingleNode;
        sptr = sptr->next;
        sptr->value = i;
        sptr->next = NULL;

        dptr->next = new DoubleNode;
        dptr->next->pre = dptr;
        dptr = dptr->next;
        dptr->value = i;
        dptr->next = NULL;
    }
    cout << "SingleNode before reversed: " << endl;
    printList(shead);
    cout << "SingleNode after reversed: " << endl;
    shead = reverseList(shead);
    printList(shead);    
    
    cout << "DoubleNode before reversed: " << endl;
    printList(dhead);
    cout << "DoubleNode after reversed: " << endl;
    dhead = reverseList(dhead);
    printList(dhead);
    return 0;
}