c - 在给定值之后反转链表，而不创建新节点

Question

我真的很坚持这个练习。我可以毫无问题地反转链表，但我不知道如何在不创建新节点的情况下做到这一点。

实现函数reverse()来反转链表中的元素从给定值第一次出现开始。例如给定输入 A B C D E F 和值 D，返回列表 A B C F E D

您应该就地执行此操作，而不创建新节点。

struct Node
{
    struct Node* next;
    int val;
};

void reverse( struct Node* head, int val);

你能给我一些建议吗？

这就是我现在所拥有的:

struct Node
{
    struct Node* next;
    int val;
};

void reverse(struct Node* head)
{
   struct Node *p = head,
                *q = NULL,
                *r;
    while (p != NULL)
    {
        r = q;
        q = p;
        p = p->next;
        q->next = r;
    }
    q = head;
}

void reverseAfter( struct Node* head, int val )
{
    while(head)
    {
        if(head->val == val)
            reverse(head);
        else
            head = head->next;  
    }
}

Answer 1

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

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

void reverse(Node **head){
    Node *tmp, *list, *newList = NULL;
    if (head==NULL || *head == NULL) return;
    list = *head;
    while(list != NULL){
        tmp = list;
        list = list->next;
        tmp->next = newList;
        newList = tmp;
    }
    *head = newList;
}

void print(Node *head){
    while(head){
        printf("%d ", head->val);
        head = head->next;
    }
    printf("\n");
}

void reverseAfter(Node *head, int val){
    //The first element excluded
    if(head == NULL) return ;

    while(head->next && head->next->val != val)
        head = head->next;
    reverse(&head->next);
}

int main(void){
    Node *head;
    Node n[6] = {
        {1,NULL}, {2, NULL}, {3, NULL},
        {4,NULL}, {5, NULL}, {6, NULL}
    };
    n[0].next = &n[1];n[1].next = &n[2];n[2].next = &n[3];
    n[3].next = &n[4];n[4].next = &n[5];

    head = &n[0];
    print(head);//1 2 3 4 5 6
    //reverse(&head);
    reverseAfter(head, 4);
    print(head);//1 2 3 6 5 4
    return 0;
}