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.
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.
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode reverseBetween(ListNode head, int m, int n) { 11 if (m == n || head == null) 12 return head; 13 ListNode dummy = new ListNode(0); 14 dummy.next = head; 15 ListNode pre = dummy, start = dummy, then = dummy, tmp = dummy; 16 // Move pre to (m - 1) position 17 for (int i = 0; i < (m - 1); i++) { 18 pre = pre.next; 19 } 20 // Start = m 21 start = pre.next; 22 then = start.next; 23 start.next = null; 24 // Move (n - m) steps for reverse 25 for (int i = 0; i < (n - m); i++) { 26 tmp = then.next; 27 then.next = start; 28 start = then; 29 then = tmp; 30 } 31 pre.next.next = then; 32 pre.next = start; 33 return dummy.next; 34 } 35 }
原文：http://www.cnblogs.com/ireneyanglan/p/4918863.html