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)
{
if(head==NULL || m==n) return head;
ListNode* p0;
ListNode* plast;
ListNode* p1;
if(m==1)
{
p0=NULL;
plast=NULL;
p1=head;
}
else
{
p0=head;
for(int i=1;i<m-1;i++)
p0=p0->next;
p1=p0->next;
plast=p0;
}
ListNode* pm=p1;
ListNode* p2;
for(int i=m;i<=n;i++)
{
p2=p1->next;
p1->next=plast;
plast=p1;
p1=p2;
}
if(m==1)
{
head=plast;
}
else
{
p0->next=plast;
}
pm->next=p1;
return head;
}
};
* 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)
{
if(head==NULL || m==n) return head;
ListNode* p0;
ListNode* plast;
ListNode* p1;
if(m==1)
{
p0=NULL;
plast=NULL;
p1=head;
}
else
{
p0=head;
for(int i=1;i<m-1;i++)
p0=p0->next;
p1=p0->next;
plast=p0;
}
ListNode* pm=p1;
ListNode* p2;
for(int i=m;i<=n;i++)
{
p2=p1->next;
p1->next=plast;
plast=p1;
p1=p2;
}
if(m==1)
{
head=plast;
}
else
{
p0->next=plast;
}
pm->next=p1;
return head;
}
};