链表逆置

代码：

#include "stdafx.h"
#include "assert.h"

typedef int DataType;

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

node * create_list(int a[], int n)
{
    if (n <= 0)
        return NULL;
    node * first = new node;
    first->data = a[0];
    first->next = NULL;
    node *cur = first;
    for (int i=1;i<n;i++)
    {
        node * next = new node;
        next->data = a[i];
        next->next = NULL;
        cur->next = next;
        cur = next;
    }
    return first;
}

void printf_list(node *head)
{
    while(head)
    {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("
");
}

node * reserve_list(node *head)
{
    //assert(head != NULL);    
    if(head == NULL)           //不管是笔试还是面试，这些细节都是考察要点！！！
        return head;
    node *pre = head, *post = head->next;
    while(post)
    {
        node * temp = post->next;
        post->next = pre;        
        pre = post;
        post = temp;
    }
    head->next = NULL;
    return pre;
}

node * recursion_reserve(node *head) //递归（也可以使用二级指针将newHead带出递归）
{
    if (head == NULL)
        return NULL;
    
    if (head->next == NULL)
        return head;
    
    node * pre = head, * post = head->next;
    
    if(post->next == NULL)
    {
        post->next = pre;
        pre->next = NULL; //去掉这句，当链表元素个数为2时，死循环。
        return post;
    }
    else
    {
        node * newHead = recursion_reserve(post);    
        post->next = pre;
        pre->next = NULL;
        return newHead;
    }        
}

int _tmain(int argc, _TCHAR* argv[])
{
    int a[] = {1,2,3,4,5,6,7,8,9};

    node * head = create_list(a, sizeof(a)/sizeof(int));
    printf_list(head);    

    node *reservedHead = reserve_list(head); //reserve_list(NULL) 
    printf_list(reservedHead);    

    node *recursionHead = recursion_reserve(reservedHead);
    printf_list(recursionHead);

    getchar();
    return 0;
}

单链表的逆置-C++实现

单链表逆置