C-我正在尝试编写一个函数，该函数通过参考逆转链接列表

Question

我不确定我的写作是否正确。我的策略首先是按原点列表的第一个节点获得的，然后创建一个新的节点的新列表(在将原始列表的下一个节点放在head节点上(，然后迭代地每次获得第一个节点和链接。它是作为该列表的负责人到新的，相反的列表。这是我到目前为止所做的：

typedef struct node {
    int data;
    struct node *next;
} Node;
void reverseList(Node **head) {
    Node *curr = *head;        //
    Node *new_node = *head;    
    Node *prev = NULL;
    new_node->next = NULL;  //the new list is to end with the first node of the origin list//
    while (curr != NULL) {      //traverse through the whole list//
        curr = curr->next;
        prev = curr;            //getting the next first node// 
        prev->next = new_node;  //and making it linked to the new list//
    }
    *head = new_node;   //the new, reversed list//
}

Answer 1

您的代码中存在逻辑错误 -
观察代码段：

Node* new_node=*head;    
Node* prev=NULL;
new_node->next=NULL;

第一行将new_node设置为head，而最后一行将next的new_node指针设置为NULL。因此，有效地将head->next设置为NULL。结果，while循环最多一次。

在这里，我给您略微修改了reverseList功能

void reverseList(Node** head)  
{
    Node* curr=*head; 
    Node *temp;       
    Node* prev=NULL;
    while(curr!=NULL)     
    {
        temp = curr->next; // keep track of the current nodes next node.
        curr->next = prev; // reverse the link for the current node
        prev=curr; // current  node becomes the previous node for next iteration
        curr=temp; // now the initially next node becomes the current node for next iteration
    }
    /*
        After the end of the whiie loop, prev points to the last node.
        So the change *head to point to the last node.
    */
    *head = prev; 
}