Link:http://oj.leetcode.com/problems/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.
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode reverseBetween(ListNode head, int m, int n) { 14 if (head == null || head.next == null) 15 return head; 16 ListNode result = new ListNode(0); 17 result.next = head; 18 head = result; 19 ListNode pre = head, cur = head, post = head; 20 int index = 1; 21 22 while (index < m) { 23 pre = pre.next; 24 index++; 25 } 26 cur = pre.next; 27 post = cur.next; 28 //in this moment, index = m. 29 //example, for{1,2,3,4}, and m = 1, n = 4, 30 //pre is the "virtual" list node point to 1, and cur point to 1, post points to 2 31 //during the iteration: 32 //2->1->3->4 33 //3->2->1->4 34 //4->3->2->1 35 while (index < n) { 36 index++; 37 //get the next element of post 38 ListNode temp = post.next; 39 //post.next should point to the next element of pre 40 post.next = pre.next; 41 //pre.next should point to the post now 42 pre.next = post; 43 //cur.next will points to the temp, and finally, cur will point to null 44 cur.next = temp; 45 //move post to temp 46 post = temp; 47 } 48 return head.next; 49 } 50 }