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.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public void reverseBetween(ListNode first,ListNode end) { if(first==end) { return; } ListNode prior=first; ListNode p=prior.next; for(;p!=end;) { ListNode next=p.next; p.next=prior; prior=p; p=next; } end.next=prior; } public ListNode reverseBetween(ListNode head, int m, int n) { if(head==null||m==n) { return head; } ListNode pHead=new ListNode(0); pHead.next=head; ListNode prior=pHead; ListNode next; ListNode first; ListNode end; int i=0; for(i=1;i<m;i++) { prior=prior.next; } first=prior.next; for(i=m,end=first;i<n;i++) { end=end.next; } next=end.next; reverseBetween(first,end); prior.next=end; first.next=next; return pHead.next; } }
原文地址:http://blog.csdn.net/jiewuyou/article/details/39178569