92. 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.

#include<iostream>
using namespace std;
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(m==n)
return head;
ListNode *p,*q,*t,*s;
ListNode *front=new ListNode(-1);
front->next=head;
p=front;
int count=1;
while(count<m)//先找到第m个结点的前驱结点
{
p=p->next;
count++;
}
q=p->next;//第m个结点保持不变
count=m;
while(count<n)
{
t=q->next;//取出q后面的元素插入到前面
q->next=t->next;//保证后面链表连接
s=p->next;//取出要插入结点后面的链表
p->next=t;//插入
t->next=s;//连接后面的链表
count++;
}
return front->next;
}
};
int main()
{
ListNode *head,*p;
head=NULL;
p=head;
int N,i;
cin>>N;
for(i=0;i<N;i++)
{
ListNode *q=(ListNode *)malloc(sizeof(ListNode));
cin>>q->val;
q->next=NULL;
if(head==NULL)
head=q;
else
p->next=q;
p=q;
}
int m,n;
cin>>m>>n;
Solution solve;
ListNode *ret=solve.reverseBetween(head,m,n);
while(ret)
{
cout<<ret->val<<' ';
ret=ret->next;
}
return 0;
}