Question

我想在C中编写一个函数来通过递归反转链表，但是我看到很多方法作为迭代方式（与我一起完成）..但是以递归方式，函数总是返回0

撤消链接列表代码

node * r_reverseItem(node * head) {    
    if (head == NULL) //only one elem
        return NULL;

    node * currentNode = head;

    if(currentNode->next == NULL) { 
        //set HEAD to current TAIL since we are reversing list
        head = currentNode; 
        return head; //since this is the base case
    }

    r_reverseItem(currentNode->next);
    currentNode->next->next = currentNode;
    currentNode->next = NULL; //set "old" next pointer to NULL
}

那么，请问这个代码中的问题是什么？但是，任何有用的例子或网站也是可取的。

编辑

只是我知道我的错在哪里。我忘了添加一个返回声明（杀手错误）。所以正确的功能如下：

   node * r_reverseItem(node * head) {    
    if (head == NULL) //only one elem
        return NULL;

    node * currentNode = head;

    if(currentNode->next == NULL) { 
        //set HEAD to current TAIL since we are reversing list
        head = currentNode; 
        return head; //since this is the base case
    }

       node * newNode =r_reverseItem(currentNode->next);
           currentNode->next->next = currentNode;
           currentNode->next = NULL; //set "old" next pointer to NULL
           return newNode;
}

Answer 1

原始解决方案implemented in Java，我只使用您的变量名称

将其翻译为C

node* r_reverseItem(node * head)
{
    if( !head )   // empty list
        return NULL;

    if( !(head->next) ) //last element becomes the head of the reversed list
        return head;

    node *currentNode = head->next; //preserve next element
    head->next = NULL;

        node* newHead = r_reverseItem(currentNode);

    currentNode->next = head; 
    return newHead; //once you have head you don't change it 
}

Answer 2

检查这个网站，基本上它解释了一个反向链接列表的三种方式，在你阅读它们之后你可以按照你想要的方式递归它

http://www.codeproject.com/Articles/27742/How-To-Reverse-a-Linked-List-Different-Ways

反转链表递归解-C

2 个答案: