[LeetCode] Reverse Linked List II

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.
最近写代码的时候老忘记在while循环里对变量做 减减或者加加。。。然后就死循环了。。

#define LN ListNode
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;
n=n-m+1;
m--;
LN guard(-1);
guard.next=head;
LN* tail1=&guard;
LN* cur=head;
while(m&&cur)
{
tail1=tail1->next;
cur=cur->next;
m--;
}
assert(cur);
LN* tail2=cur;
tail1->next=NULL;
while(cur&&n)
{
LN* tmp=cur->next;
cur->next=tail1->next;
tail1->next=cur;
cur=tmp;
n--;
}
assert(n==0);
tail2->next=cur;
return guard.next;
}
};