https://leetcode.com/problems/reverse-linked-list/description/
Hint:
A linked list can be reversed either iteratively or recursively. Could you implement both?
test case:      3->1->2>5 to 5->2->1->3
           x    c

 1 public class LC206_ReverseLinkedList {
 2     public ListNode reverseLinkedList(ListNode head){
 
 3         if (head == null || head.next == null) return head ;
 4         ListNode prev = null, curr= head, next = null ;
 5         /*
 6         * 細節， 如果 寫成curr.next != null, curr 會在 NODE4 但是沒有進入循環，所以並不會連上前面的點，
 7         * 所以返回的時候會是一個單點，而不是鏈表
 8         * thats the reason curr should be stopped at null to make sure all prev nodes linked.
 
 9         * */
10         while (curr != null ){
11             next = curr.next ;
12             curr.next = prev ;
13             prev = curr ;
14             curr = next ;
15         }
16         return prev ;
17     }
18 
19     public ListNode reverseLinkedList2(ListNode head){
20         //base
21
 
         /*
22         * 1-2-3-4   null<-1<-2<-3<-4
23         * current iteration head =2  expect null<-3<-4
24         *
25         * */
26         ListNode newHead = reverseLinkedList2(head.next) ;
27         ListNode tail = head.next ; //3
28         tail.next  = head ; //2<-3<-4
29         head.next = null;  //null<-2<-3<-4
30         return newHead ;
31     }
32 }

recursive: assume Node2 is the head, subproblem return:

null<-Node3 <- Node4　　

then you want to change to

ListNode newHead = reverse(head.next) will generate the following pic:

ListNode tail = head.next ; here tail is Node 3

note here since we didnt touch head, thats the reason node2 still points to node3

tail.next = head

head.next = null; //this is what you need to return for the upper level recursive

因為這裏是recursive, 一層層走下去 reverse(head.next) 所以base case:

if(head == null || head.next == null) 最後head 會是 原 linkedlist 的尾巴，也就是新頭。

LC_206. Reverse Linked List