iterative: Core code
if(head == null || head.next == null) return head; ListNode current = head.next; head.next = null; while(current ! = null){ ListNode temp = current.next; current.next = head; head = current; current = temp.next; } return head;
recurring : Core code
public ListNode reverseList(ListNode head) { if(head == null || head.next == null) return head; ListNode second = head.next; head.next = null; ListNode res = reverseList(second); second.next = head; return res; }
Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL
, m = 2 and n = 4,
return 1->4->3->2->5->NULL
.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
two point: reverse
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode reverseBetween(ListNode head, int m, int n) { if(head == null || head.next == null) return head; ListNode res = new ListNode(-1); res.next = head; ListNode pre = res; for(int i = 0; i < m - 1 ; i ++){ pre = pre.next; } ListNode first = pre.next; ListNode second = first.next; for(int i = 0; i < n - m ; i++){ first.next = second.next; second.next = pre.next; pre.next = second; second = first.next; } return res.next; } }