[leetcode]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.

我给这个题深深的跪了。。。

/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(m == n) return head;

ListNode *p = NULL, *q = head;

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

ListNode *e1 = p, *s1 = q;

p = q;
q = q->next;

ListNode *r;
for(int i = m; i < n; i++){
r = q->next;
q->next = p;
p = q;
q = r;
}

if(s1) s1->next = q;

if(e1) e1->next = p;
else head = p;

return head;

}
};