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.
class Solution {
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
if(head == NULL)
return NULL;
ListNode dummy(0);
dummy.next = head;
ListNode *pre = &dummy;
ListNode *p = head;
int k = 1;
while(k < m)
{
pre = p;
p = p->next;
k++;
}
ListNode *tail = NULL;
ListNode *nxt = NULL;
ListNode *tmp = p;
while(k <= n)
{
nxt = p->next;
p->next = tail;
tail = p;
p = nxt;
k++;
}
pre->next = tail;
tmp->next = p;
return dummy.next;
}
};Leetcode:Reverse Linked List II 单链表区间范围内逆置,布布扣,bubuko.com
Leetcode:Reverse Linked List II 单链表区间范围内逆置
原文地址：http://blog.csdn.net/u012118523/article/details/26407127
