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反转链表

[英]Reversing a linked list

My problem is: Given a function to reverse a linked list. 我的问题是:给定一个反转链表的功能。

My attempt at it in C was: 我在C中的尝试是:

ListNode *reverse(ListNode *head)
{
    if(head == NULL || head->next == NULL)
        return head;

    ListNode *temp = head->next;
    ListNode *retP =  reverse(temp);
    temp->next = head;
    head->next = NULL;
    return retP;
}

But I do not think this is right. 但我不认为这是对的。 I want to be able to do it in Java and I am stumped on this. 我希望能够用Java做到这一点,我对此感到难过。 Any help would be appreciated. 任何帮助,将不胜感激。 Please help me get started 请帮我开始吧

If you want to reverse a List in Java, use 如果要在Java中反转List,请使用

Collections.reverse(List list)

If you want to know how it is implemented or want to do it by hand, have a look at the JDK sources of java.util.Collections . 如果您想知道它是如何实现的或想要手动完成,请查看java.util.Collections的JDK源代码

Iteratively 迭代

public reverseListIteratively (Node head) {
if (head == NULL || head.next == NULL)
return;  //empty or just one node in list

Node Second = head.next;

//store third node before we change 
Node Third = Second.next;  

//Second's next pointer
Second.next = head;  //second now points to head
head.next = NULL;  //change head pointer to NULL

//only two nodes, which we already reversed
if (Third == NULL)
return;  

Node CurrentNode = Third;

Node PreviousNode = Second;

while (CurrentNode != NULL)
{ 
Node NextNode = CurrentNode.next;

CurrentNode.next = PreviousNode;

/*  repeat the process, but have to reset
     the PreviousNode and CurrentNode
*/

PreviousNode = CurrentNode;
CurrentNode = NextNode;  
}

head  = PreviousNode; //reset the head node
}

Recursively 递归

public void recursiveReverse(Node currentNode )
{  
 //check for empty list 
 if(currentNode == NULL)
    return;

/* if we are at the TAIL node:
    recursive base case:
 */
if(currentNode.next == NULL) 
{ 
//set HEAD to current TAIL since we are reversing list
head = currentNode; 
return; //since this is the base case
}

recursiveReverse(currentNode.next);
currentNode.next.next = currentNode;
currentNode.next = null; //set "old" next pointer to NULL
}

Source, with explanation (after using google for 3 seconds) http://www.programmerinterview.com/index.php/data-structures/reverse-a-linked-list/ 来源,有解释(使用谷歌3秒后) http://www.programmerinterview.com/index.php/data-structures/reverse-a-linked-list/

public Node reverse(Node node){
    Node p=null, c=node, n=node;
    while(c!=null){
        n=c.next;
        c.next=p;
        p=c;
        c=n;
    }
return p;
}

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