Reverse Linked List II 解答

Question

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.

Solution

Four pointers:

Dummy node, pre, start, then, tmp

First, we find the position to start reversal.

Key point here is to use Dummy node, and pre points the (m - 1) position.

Then, we use three pointers, start, then, tmp to implement reversal. Just the same way as Reverse Linked List.

Several points to note here:

1. Move (m - 1) steps to get pre position

2. Move (n - m) steps to implement reversal

3. Link reversed sub-list with original unreversed sub-lists.

In-place and in one-pass.

 /**
  * 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 (m == n || head == null)
             return head;
         ListNode dummy = new ListNode(0);
         dummy.next = head;
         ListNode pre = dummy, start = dummy, then = dummy, tmp = dummy;
         // Move pre to (m - 1) position
         for (int i = 0; i < (m - 1); i++) {
             pre = pre.next;
         }
         // Start = m
         start = pre.next;
         then = start.next;
         start.next = null;
         // Move (n - m) steps for reverse
         for (int i = 0; i < (n - m); i++) {
             tmp = then.next;
             then.next = start;
             start = then;
             then = tmp;
         }
         pre.next.next = then;
         pre.next = start;
         return dummy.next;
     }
 }