LC_206. Reverse Linked List

 

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

public class LC206_ReverseLinkedList {
    public ListNode reverseLinkedList(ListNode head){
        if (head == null || head.next == null) return head ;
        ListNode prev = null, curr= head, next = null ;
        /*
        * 细节， 如果 写成curr.next != null, curr 会在 NODE4 但是没有进入循环，所以并不会连上前面的点，
        * 所以返回的时候会是一个单点，而不是链表
        * thats the reason curr should be stopped at null to make sure all prev nodes linked.
        * */
        while (curr != null ){
            next = curr.next ;
            curr.next = prev ;
            prev = curr ;
            curr = next ;
        }
        return prev ;
    }

    public ListNode reverseLinkedList2(ListNode head){
        //base
　　　　　　if (head == null || head.next == null) return head ;

        /*
        * 1-2-3-4   null<-1<-2<-3<-4
        * current iteration head =2  expect null<-3<-4
        *
        * */
        ListNode newHead = reverseLinkedList2(head.next) ;
        ListNode tail = head.next ; //3
        tail.next  = head ; //2<-3<-4
        head.next = null;  //null<-2<-3<-4
        return newHead ;
    }
}

 

 

 

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

 

因为这里是recursive, 一层层走下去 reverse(head.next) 所以base case:

if(head == null || head.next == null) 最后head 会是 原 linkedlist 的尾巴，也就是新头。