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在 C 中反轉鏈表

[英]Reversing Linked List in C

我一直在研究一個代碼來反轉一個鏈表,但我似乎只是在我的列表中打印出第一個元素。 這是我的功能:

void reverseSqueue(Squeue squeue) {
    if (squeue -> first == NULL || squeue -> last == NULL){
        return;
    }
    Node* temp = NULL;
    Node* curr = squeue -> first;
    squeue -> first = squeue -> last;
    squeue -> last = curr;

    /*if (squeue -> first != squeue -> last){
      curr = curr -> next;
    }

    printf("%s\n", curr -> next -> value);
    printf("%s\n", curr -> value);
    printf("%s\n", curr -> prev -> value);
    printf("______________\n");
    printf("%s\n", squeue -> first -> value);
    Node* nde = squeue -> first;
    while (nde != NULL){
        printf("%s\n", nde -> value);
    nde = nde -> prev;
    }*/
    while (curr != NULL){
    /*if (curr != squeue -> first || curr != squeue -> last){
        if (curr -> next == squeue -> last){
            curr = NULL;
        }else{*/
        temp = curr;
        curr -> next = curr;
        curr = temp;
        curr = curr -> prev;
      //}
    //}else{
    //curr = NULL;
    //}
    }
}

我為混亂的代碼道歉,我嘗試運行多個測試並在此處注釋掉它們; 我之前的代碼運行良好,但在 OClint 中給出了垃圾值警告,所以我嘗試修復它們,現在我的代碼似乎不再起作用了。 如果沒有其他工作,我可以回到我的原始代碼。 非常感謝任何輸入!

struct node {
  char *value;
  struct node *next;
  struct node *prev;
};
typedef struct node Node;

struct squeue {
   struct node* first;
   struct node* last;
};
typedef struct squeue * Squeue;

使用 curr->next = curr 你是說當前節點的下一個節點是當前節點,也就是說,當前節點指向自己,因為它是下一個節點。

Node *first = squeue->last; 
squeue->last = squeue->first; // the last node is now the first
squeue->first = first; // squeue->first = original order squeue->last

Node *curr = first; 
while (curr != NULL) { // loop begins with the original order last node,
                       // but new order first node

    // curr->next, curr->prev = curr->prev, curr->next
    Node *prev = curr->prev; // saves the curr's prev
    curr->prev = curr->next; // modifies curr's prev, now pointing to curr's next
    curr->next = prev; // curr's next now points to it's original prev

    curr = curr->next; // the next curr is the original order curr's prev,
                       // but the now order curr's next
}

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