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-

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:

ListNode *reverse(ListNode * head ,ListNode *start,ListNode * end)
{
ListNode *tmp,*p,*q;

tmp = head;
p=q=NULL;

if( start != head )
{

while(tmp->next != start)
{
tmp = tmp->next;
}

p=tmp;

}

q=end->next;
end->next = NULL;

end=start;
tmp=start->next;
start->next = NULL;
ListNode * tp;

while( tmp != NULL )
{
tp = tmp;
tmp= tmp->next;
tp->next=start;
start = tp;
}

if(p==NULL)
{
head = start;
}

if(p !=NULL)
{
p->next=start;

}

if(q !=NULL)
{
end->next = q;
}
return head;
}

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

ListNode * p, *q;
p=q=head;
int tmp;

if( m < n )
{
for( int i = 1; i < m && p->next != NULL ; ++i )
{
p=p->next;

}

q=p;

for( int i = m; i < n && q->next != NULL; ++i )
{
q=q->next;
}

if( 1== n-m || 2== n-m )
{
tmp=p->val;
p->val = q->val;
q->val = tmp;
}
else
{
return reverse( head,p,q);
}

}

return head;
}
};