c - 如何在 C 中反转单链表？

Question

对于我的类(class)，我的任务是构建链表、输入数据和执行不同的功能。我的代码编译正常，但是还原和建设性还原功能不起作用。

如果我反转列表并打印它，我只会得到第一个节点。我似乎在某处错过了重点。这是我的代码:(如果您发现任何其他功能有任何错误，请告诉我)提前致谢!

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

typedef struct DoubleNode{
    struct DoubleNode* next;
    double data;

} DoubleNode;

DoubleNode* insertFirst(DoubleNode*head, double c) {
    DoubleNode* temp;
    temp=malloc(sizeof(DoubleNode));
    temp->data= c;
    temp->next= head;

    return temp;
}
void printList(DoubleNode*head) {
    DoubleNode* cursor;
    printf("(");
    cursor=head;
    while (cursor != NULL) {
        printf("%fl", cursor->data);
        printf(" ");
        cursor=cursor->next;
    }
    printf(")");
}


DoubleNode* insertLast(DoubleNode *head, double d) {
    DoubleNode *tmp, *cursor;
    //trivial case: empty list
    if (head==NULL)
    {return insertFirst(head,d);
    } else{
        //general case: goto end
        cursor=head;
        while (cursor->next != NULL){
            cursor =cursor->next;
        }
        tmp= malloc(sizeof(DoubleNode));
        tmp->data = d;
        tmp->next = NULL;
        cursor->next=tmp;
        return head;
    }
}

DoubleNode* reverseDoubleListCon(DoubleNode*head) {
    DoubleNode *temp, *res, *cell;
    cell=head;
    res=NULL;
    while (cell!=NULL) {
        temp=malloc(sizeof(DoubleNode));
        temp->data=cell->data;
        temp->next=res;
        res=temp;
        cell=cell->next;
    }
    return res;
}

void reverseDoubleList(DoubleNode*head) {
    DoubleNode *chain, *revChain, *cell;
    cell=head;
    revChain=NULL;
    while (cell!=NULL) {
        chain=cell->next;
        cell->next=revChain;
        revChain=cell;
        cell=chain;
    }
    head=revChain;
}

double get(DoubleNode*head, double n) {
    DoubleNode *cursor;
    cursor=head;
    for (int i=0; i<n; i++) {
        cursor=cursor->next;

    }
    return cursor->data;
}

DoubleNode* delete(DoubleNode*head, double n) {
    DoubleNode *cursor, *rest;
    cursor=head;
    for (int i=0; i<n; i++) {
        cursor=rest;
        cursor=cursor->next;

    }
    rest->next=cursor->next;
    free(cursor);
    return head;

}

DoubleNode* insert(DoubleNode*head, double d, double n) {
    DoubleNode *cursor, *temp, *rest;
    cursor=head;

    for (int i = 0; i<n-1; i++) {
        rest=cursor;
        cursor= cursor->next;
    }
    temp=malloc(sizeof(DoubleNode));
    temp->data=d;
    temp->next=cursor;
    rest->next= temp;

    return head;
}

int main(int argc, const char * argv[]) {
    DoubleNode *head;
    head=malloc(sizeof(DoubleNode));

    for (double i=0.0; i <11.0; i++) {
        head=insertLast(head, i);
    }
    printList(head);
    reverseDoubleList(head);
    printList(head);
    reverseDoubleListCon(head);
    printList(head);

    return 0;
}

Answer 1

void rev(node **head)
{
    node *prev = NULL;
    node *next = *head->next;
    node *cur;

    for (cur = *head; cur; cur = next) {
        next = cur->next;
        prev = cur;
    }
    *head = prev
}

首先是链表:

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第二步:让节点B指向A，让A->next为NULL

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第 3 步:该过程继续......

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第 4 步:链表现在反转了。