单链表逆转(自己根据题干实现)

链接

题目:

PTA成功代码:

List Reverse( List L )
{
    List x, y;
    x = L;
    L = NULL;
    while (x)
    {
        y = x;
        x = x->Next;
        y->Next = L;
        L = y;
    }
    return L;
}

本地代码(实现了输入逆转输出释放):

//https://pintia.cn/problem-sets/15/problems/724
//链表倒转

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
#include <math.h>

// 定义一个结构体
typedef struct student
{
    int ID;
    struct student *next;

} LinkList;

// 创建首尾节点和普通节点
LinkList *head = NULL, *node = NULL, *end = NULL;
// 创建一个链表,链表节点个数为n
LinkList *create( int n )
{
    for (int i = 0; i < n; i++)
    {
        node = ( LinkList* )malloc( sizeof( LinkList ) );
        memset(node,0,sizeof(node));
        printf("请输入第%d个人的ID\n", i+1);
        scanf("%d",&node->ID);
        if (i == 0)
        {
            head = node;
            end = head;
        }
        end->next = node;
        end = node;
    }
    end->next = NULL;
    return head;
}

// 循环遍历然后打印
void put( LinkList *head, int n )
{
    LinkList *t = head;
    while (1)
    {
        printf("%d\n", t->ID );
        t = t->next;
        if ( t->next == NULL )
        {
            printf("%d\n", t->ID );
            break;
        }
    }
}

// 把链表倒叙
LinkList *reverse( LinkList *head )
{
    LinkList *x, *y;
    x = head;
    head = NULL;
    while ( x != NULL )
    {
        y = x;
        x = x->next;
        y->next = head;
        head = y; 
    }
    return head;
}

// 释放内存
void freeLinkList( LinkList *head )
{
    LinkList *freeNode = NULL;
    while ( head != NULL)
    {
        freeNode = head;
        head = head->next;
        free( freeNode );
    }
}

int main()
{
    int n;
    scanf("%d", &n);
    LinkList *head;
    head = create(n);
    put( head, n );
    puts("After reverse:");
    head = reverse( head );
    put( head, n );

    freeLinkList( head );

    return 0;
}

单链表逆转思路:

链接