实验11-2-9 链表逆置 (20 分)
1. 题目摘自
https://pintia.cn/problem-sets/13/problems/607
2. 题目内容
本题要求实现一个函数，将给定单向链表逆置，即表头置为表尾，表尾置为表头。链表结点定义如下：
struct ListNode {
int data;
struct ListNode *next;
};
函数接口定义：
struct ListNode *reverse( struct ListNode *head );
其中head是用户传入的链表的头指针；函数reverse将链表head逆置，并返回结果链表的头指针。
输入样例：
1 2 3 4 5 6 -1
输出样例：
6 5 4 3 2 1
3. 源码参考
#include <iostream>
#include <stdlib.h>
using namespace std;
struct ListNode {
int data;
struct ListNode *next;
};
struct ListNode *createlist(); /*裁判实现，细节不表*/
struct ListNode *reverse( struct ListNode *head );
void printlist( struct ListNode *head )
{
struct ListNode *p = head;
while (p) {
printf("%d ", p->data);
p = p->next;
}
printf("\n");
}
int main()
{
struct ListNode *head;
head = createlist();
head = reverse(head);
printlist(head);
return 0;
}
struct ListNode *createlist()
{
struct ListNode *p, *h, *t;
int n;
h = NULL;
cin >> n;
while(n != -1)
{
p = (struct ListNode*)malloc(sizeof(struct ListNode));
p->data = n;
p->next = NULL;
if(h == NULL)
{
h = p;
}
else
{
t->next = p;
}
t = p;
cin >> n;
}
return h;
}
struct ListNode *reverse( struct ListNode *head )
{
struct ListNode *pPre, *pCur, *pNext;
pPre = head;
pCur = pPre->next;
pNext = NULL;
while(pCur)
{
pNext = pCur->next;
pCur->next = pPre;
pPre = pCur;
pCur = pNext;
}
head->next = NULL;
head = pPre; //记录下新的头结点
return head;
}
