Leetcode139: Reverse Linked List II

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.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
if (head == NULL)
return NULL;

ListNode *q = NULL;
ListNode *p = head;
for(int i = 0; i < m - 1; i++)
{
q = p;
p = p->next;
}

ListNode *end = p;
ListNode *pPre = p;
p = p->next;
for(int i = m + 1; i <= n; i++)
{
ListNode *pNext = p->next;

p->next = pPre;
pPre = p;
p = pNext;
}

end->next = p;
if (q)
q->next = pPre;
else
head = pPre;

return head;

}
};