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.
Analysis:
We need three pointers: pre, cur and nextCur. This is because when we reverse cur.next to pre, we lose the link between cur and cur.next, then we cannot go on. We need record the node after cur also.
We also need to address the case that m==1 carefully. In this case, the head of the list is changed.
Solution:
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) return head; 15 if (m==n) return head; 16 17 ListNode reHead = null; 18 ListNode reHeadPre = null; 19 ListNode cur=null; 20 ListNode pre = null; 21 int index = 1; 22 cur = head; 23 while (index!=m){ 24 pre = cur; 25 cur = cur.next; 26 index++; 27 } 28 29 reHead = cur; 30 reHeadPre = pre; 31 ListNode nextCur = cur.next; 32 33 while (index!=n){ 34 pre = cur; 35 //NOTE: it is not "cur = cur.next;", because the cur.next has been changed to pre! 36 cur = nextCur; 37 nextCur = nextCur.next; 38 index++; 39 cur.next = pre; 40 } 41 42 //Address the case that m==1. 43 if (reHeadPre==null) 44 head = cur; 45 else 46 reHeadPre.next=cur; 47 reHead.next=nextCur; 48 49 return head; 50 } 51 }