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) {
int d=n-m;
if(head==NULL||head->next==NULL||d<=)
return head; ListNode* p=head,*bfp=p;
ListNode* q=head;
for(int i=;i<m-;i++){
bfp=p;
p=p->next;
}
ListNode* revhead=p;
ListNode* bf=p,*bh=NULL;
p=p->next;
d--;
while(p->next!=NULL&&d!=){
d--;
bh=p->next;
p->next=bf;
bf=p;
p=bh; }
bh=p->next;
p->next=bf;
bfp->next=p;
revhead->next=bh;
if(m!=)
return head;
else return p;
}
};