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;
}
}