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.
先反转m到n之间的list再头尾衔接一下
/**
* 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) {
ListNode dummy(-1);
dummy.next=head;
ListNode* pm=&dummy;
for(int i=0;i<m-1;i++)
pm=pm->next;
ListNode* start=pm;
pm=pm->next;
ListNode* begin=pm;
ListNode *next,*cur=pm->next;
for(int i=n-m;i>0;i--)
{
next=cur->next;
cur->next=pm;
pm=cur;
cur=next;
}
start->next=pm;
begin->next=cur;
return dummy.next;
}
};