2118 数据结构实验之链表三：链表的逆置

数据结构实验之链表三：链表的逆置

Time Limit: 1000MS Memory Limit: 65536KB

Problem Description

输入多个整数，以-1作为结束标志，顺序建立一个带头结点的单链表，之后对该单链表的数据进行逆置，并输出逆置后的单链表数据。

Input

输入多个整数，以-1作为结束标志。

Output

输出逆置后的单链表数据。

Example Input

12 56 4 6 55 15 33 62 -1

Example Output

62 33 15 55 6 4 56 12

Hint

不得使用数组。

#include <bits/stdc++.h>using namespace std;typedef struct node{    int date;    struct node *next;}Lnode;Lnode *creat(int x)//建表 输入{    Lnode *p,*q,*head;    int k;    //head = (Lnode*)malloc(sizeof(Lnode));    head = new node;    head -> next = NULL;    q = head;    p = new node;    p -> next = NULL;    p->date = x;    q ->next = p;    q = p;    while(cin >> k && k != -1 )    {        p = new node;        p -> next = NULL;        p->date = k;        q ->next = p;        q = p;    }    return head;}void output( Lnode *head){    Lnode *q;    q = head ->next;    cout << q ->date;    while(q->next)    {        cout << " " << q ->next ->date;        q = q ->next;    }    cout << endl;}Lnode *change(Lnode * head)//元素逆置{    Lnode *p, *q, *L;    L = new node;    L -> next = NULL;    q = L;    while(head->next)    {        p = new node;        p -> next = NULL;        p ->date = head ->next ->date;        p ->next = q ->next;        q ->next = p;        head = head ->next;    }    return L;}int main(){    int x;    Lnode *head;    cin >> x;    head = creat(x);    head = change(head);    output(head);    return 0;}