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 * 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， 7 * 所以返回的时候会是一个单点，而不是链表 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。
