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.
ref: http://www.cnblogs.com/feiling/p/3263501.html
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode reverseBetween(ListNode head, int m, int n) { if(head == null || m == n) return head; ListNode fake = new ListNode(Integer.MIN_VALUE); fake.next = head; head = fake; int count = 0; ListNode pre = null, p = head, p1 = null, p2 = null, p3 =null; while(count <= n){ if(count == m-1){ p1 = p; }else if(count == m){ p2 = p; }else if(count == n){ p3 = p; } ListNode tmp = p.next; if(count >= m+1 && count <= n){ p.next = pre; } pre = p; p = tmp; count++; } p1.next = p3; p2.next = p; return fake.next; } }
原文:http://www.cnblogs.com/RazerLu/p/3551085.html