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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.
要求在1*n时间内完成
类似于Leetcode 206 Reverse Linked List
保存指针指向倒置链表段的前一个node，在倒置结束后连接其下一个node。
var reverseBetween = function(head, m, n) { if(m===n) return head var dummy = new ListNode() dummy.next = head var p = dummy for(var i=0;i<m-1;i++) p = p.next var c = p.next var r = null for(var i=0;i<n-m+1;i++){ var nextnode = c.next c.next = r r = c c = nextnode } p.next.next = c p.next = r return dummy.next }
Leetcode 92 Reverse Linked List II
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原文地址：http://www.cnblogs.com/lilixu/p/4605493.html
