如何使用仅两个指针反转单向链表？

如果不使用临时变量，交换两个变量的值：

a = a xor b
b = a xor b
a = a xor b

最快的方法是将其写成一行。

a = a ^ b ^ (b=a)

同样地，

使用两次交换操作

swap(a,b)
swap(b,c)

XOR解决方案

a = a^b^c
b = a^b^c
c = a^b^c
a = a^b^c

一行解决方案

c = a ^ b ^ c ^ (a=b) ^ (b=c)
b = a ^ b ^ c ^ (c=a) ^ (a=b)
a = a ^ b ^ c ^ (b=c) ^ (c=a)

typedef struct List
{
 int info;
 struct List *next;
}List;


List* reverseList(List *head)
{
 p=head;
 q=p->next;
 p->next=NULL;
 while(q)
 {
    q = (List*) ((int)p ^ (int)q ^ (int)q->next ^ (int)(q->next=p) ^ (int)(p=q));
 }
 head = p;
 return head;
}  

#include<stdio.h>
#include<conio.h>

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

typedef struct NODE node;

void reverse(node **head);
void add_end(node **head,int val);
void alloc(node **p);
void print_all(node *head);

void main()
{
    node *head;
    clrscr();
    head = NULL;
    add_end( &head, 1 );
    add_end( &head, 2 );
    add_end( &head, 3 );
    print_all( head );
    reverse( &head );
    print_all( head );
    getch();
}
void alloc(node **p)
{
    node *temp;
    temp = (node *) malloc( sizeof(node *) );
    temp->next = NULL;
    *p = temp;
}
void add_end(node **head,int val)
{
    node *temp,*new_node;
    alloc(&new_node);
    new_node->value = val;
    if( *head == NULL )
    {
        *head = new_node;
        return;
    }
    for(temp = *head;temp->next!=NULL;temp=temp->next);
    temp->next = new_node;
}
void print_all(node *head)
{
    node *temp;
    int index=0;
    printf ("\n\n");
    if (head == NULL)
    {
        printf (" List is Empty \n");
        return;
    }
    for (temp=head; temp != NULL; temp=temp->next,index++)
        printf (" %d ==> %d \n",index,temp->value);
}
void reverse(node **head)
{
    node *next,*new_head;
    new_head=NULL;
    while(*head != NULL)
    {
        next = (*head)->next;
        (*head)->next = new_head;
        new_head = (*head);
        (*head) = next;
    }
    (*head)=new_head;
}

#include <stdio.h>
#include <malloc.h>

tydef struct node
{
    int info;
    struct node *link;
} *start;

void main()
{
    rev();
}

void rev()
{
    struct node *p = start, *q = NULL, *r;
    while (p != NULL)
    {
        r = q;
        q = p;
        p = p->link;
        q->link = r;
    }

    start = q;
}

计算您现在使用的算法的时间复杂度，很明显它无法改进。

curr = head;
prev = NULL;

while (curr != NULL) {
    next = curr->next; // store current's next, since it will be overwritten
    curr->next = prev;
    prev = curr;
    curr = next;
}

head = prev; // update head

你可以通过只使用一个额外指针来解决这个问题，这个指针必须为反转函数保持静态。它的时间复杂度为O(n)。

#include<stdio.h>
#include<stdlib.h>

typedef struct List* List;
struct List {
   int val;
   List next;
};

List reverse(List list) { /* with recursion and one static variable*/
    static List tail;
    if(!list || !list->next) {
        tail = list;

        return tail;
    } else {
        reverse1(list->next);
        list->next->next = list;
        list->next = NULL;

        return tail;
    }
}

使用两个指针，同时保持时间复杂度为O(n)（最快的可达到的时间复杂度），可能只能通过指针的数字转换和交换它们的值来实现。以下是一个实现示例：

#include <stdio.h>

typedef struct node
{
    int num;
    struct node* next;
}node;

void reverse(node* head)
{
   node* ptr;
   if(!head || !head->next || !head->next->next) return;
   ptr = head->next->next;
   head->next->next = NULL;
   while(ptr)
   {
     /* Swap head->next and ptr. */
     head->next = (unsigned)(ptr =\
     (unsigned)ptr ^ (unsigned)(head->next =\
     (unsigned)head->next ^ (unsigned)ptr)) ^ (unsigned)head->next;

     /* Swap head->next->next and ptr. */
     head->next->next = (unsigned)(ptr =\
     (unsigned)ptr ^ (unsigned)(head->next->next =\
     (unsigned)head->next->next ^ (unsigned)ptr)) ^ (unsigned)head->next->next;
   }
}

void add_end(node* ptr, int n)
{
    while(ptr->next) ptr = ptr->next;
    ptr->next = malloc(sizeof(node));
    ptr->next->num = n;
    ptr->next->next = NULL;
}

void print(node* ptr)
{
    while(ptr = ptr->next) printf("%d ", ptr->num);
    putchar('\n');
}

void erase(node* ptr)
{
    node *end;
    while(ptr->next)
    {
        if(ptr->next->next) ptr = ptr->next;
        else
        {
            end = ptr->next;
            ptr->next = NULL;
            free(end);
        }
    }
}

void main()
{
    int i, n = 5;
    node* dummy_head;
    dummy_head->next = NULL;
    for(i = 1; i <= n ; ++i) add_end(dummy_head, i);
    print(dummy_head);
    reverse(dummy_head);
    print(dummy_head);
    erase(dummy_head);
}

class Node {
    Node next;
    int data;

    Node(int item) {
        data = item;
        next = null;
    }
}

public class LinkedList {

    static Node head;

    //Print LinkedList
    public static void printList(Node node){

        while(node!=null){
            System.out.print(node.data+" ");
            node = node.next;
        }
        System.out.println();
    }

    //Reverse the LinkedList Utility
    public static Node reverse(Node node){

        Node new_node = null;

        while(node!=null){

            Node next = node.next;
            node.next = new_node;
            new_node = node;
            node = next;

        }
        return new_node;
    }

    public static void main(String[] args) {

        //Creating LinkedList
        LinkedList.head = new Node(1);
        LinkedList.head.next = new Node(2);
        LinkedList.head.next.next = new Node(3);
        LinkedList.head.next.next.next = new Node(4);

        LinkedList.printList(LinkedList.head);

        Node node = LinkedList.reverse(LinkedList.head);

        LinkedList.printList(node);

    }


}

using 2-pointers....bit large but simple and efficient

void reverse()

{

int n=0;

node *temp,*temp1;

temp=strptr;

while(temp->next!=NULL)

{

n++;      //counting no. of nodes

temp=temp->next;

}
// we will exchange ist by last.....2nd by 2nd last so.on....
int i=n/2;  

temp=strptr;

for(int j=1;j<=(n-i+1);j++)

temp=temp->next;
//  i started exchanging from in between ....so we do no have to traverse list so far //again and again for exchanging

while(i>0)

{

temp1=strptr;

for(int j=1;j<=i;j++)//this loop for traversing nodes before n/2

temp1=temp1->next;

int t;

t=temp1->info;

temp1->info=temp->info;

temp->info=t;

i--;

temp=temp->next; 

//at the end after exchanging say 2 and 4 in a 5 node list....temp will be at 5 and we will traverse temp1 to ist node and exchange ....

}

}

#include <stdio.h>
#include <stdlib.h>

struct node {
    int data;
    struct node* next;
};

struct node* head = NULL;

void printList(char* msg) {
    struct node* current = head;

    printf("\n%s\n", msg);

    while (current != NULL) {
        printf("%d ", current->data);
        current = current->next;
    }
}

void insert_beginning(int data) {
    struct node* newNode = (struct node*) malloc(sizeof(struct node));

    newNode->data = data;
    newNode->next = NULL;

    if (head == NULL)
    {
        head = newNode;
    } else {
        newNode->next = head;
        head = newNode;
    }
}

void insert_at(int data, int location) {

    struct node* newNode = (struct node*) malloc(sizeof(struct node));

    newNode->data = data;
    newNode->next = NULL;

    if (head == NULL)
    {
        head = newNode;
    }

    else {
        struct node* currentNode = head;
        int index = 0;

        while (currentNode != NULL && index < (location - 1)) {
            currentNode = currentNode->next;
            index++;
        }

        if (currentNode != NULL)
        {
            if (location == 0) {
                newNode->next = currentNode;
                head = newNode;
            } else {
                newNode->next = currentNode->next;
                currentNode->next = newNode;
            }
        }
    }
}


int delete_from(int location) {

    int retValue = -1;

    if (location < 0 || head == NULL)
    {
        printf("\nList is empty or invalid index");
        return -1;
    } else {

        struct node* currentNode = head;
        int index = 0;

        while (currentNode != NULL && index < (location - 1)) {
            currentNode = currentNode->next;
            index++;
        }

        if (currentNode != NULL)
        {
            // we've reached the node just one prior to the one we want to delete

            if (location == 0) {

                if (currentNode->next == NULL)
                {
                    // this is the only node in the list
                    retValue = currentNode->data;
                    free(currentNode);
                    head = NULL;
                } else {

                    // the next node should take its place
                    struct node* nextNode = currentNode->next;
                    head = nextNode;
                    retValue = currentNode->data;
                    free(currentNode);
                }
            } // if (location == 0)
            else {
                // the next node should take its place
                struct node* nextNode = currentNode->next;
                currentNode->next = nextNode->next;

                if (nextNode != NULL
                ) {
                    retValue = nextNode->data;
                    free(nextNode);
                }
            }

        } else {
            printf("\nInvalid index");
            return -1;
        }
    }

    return retValue;
}

void another_reverse() {
    if (head == NULL)
    {
        printf("\nList is empty\n");
        return;
    } else {
        // get the tail pointer

        struct node* tailNode = head;
        int index = 0, counter = 0;

        while (tailNode->next != NULL) {
            tailNode = tailNode->next;
            index++;
        }

        // now tailNode points to the last node
        while (counter != index) {
            int data = delete_from(index);
            insert_at(data, counter);
            counter++;
        }
    }
}

int main(int argc, char** argv) {

    insert_beginning(4);
    insert_beginning(3);
    insert_beginning(2);
    insert_beginning(1);
    insert_beginning(0);

    /*  insert_at(5, 0);
     insert_at(4, 1);
     insert_at(3, 2);
     insert_at(1, 1);*/

    printList("Original List\0");

    //reverse_list();
    another_reverse();

    printList("Reversed List\0");

    /*  delete_from(2);
     delete_from(2);*/

    //printList();
    return 0;
}