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.

 

C++实现代码：

#include<iostream>

#include<new>

using namespace std;



//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)

            return NULL;

        ListNode *p=head;

        ListNode *pre=head;

        ListNode *qpre=NULL;

        ListNode *q=NULL;

        int count=0;

        while(p)

        {

            count++;

            if(count==m)

            {

                qpre=pre;

                q=p;

            }

            else if(count==n)

                break;

            pre=p;

            p=p->next;

        }

        cout<<p->val<<endl;

        cout<<q->val<<endl;

        if(p==NULL||p==q)

            return head;

        ListNode *pp=p->next;

        ListNode *qq=NULL;

        p->next=NULL;

        p=q->next;

        cout<<p->val<<endl;

        q->next=pp;

        while(p)

        {

            qq=p->next;

            cout<<p->val<<endl;

            p->next=q;

            q=p;

            cout<<q->val<<endl;

            p=qq;

        }

        cout<<q->val<<endl;

        if(m!=1)

            qpre->next=q;

        else

            head=q;

        return head;

    }

    void createList(ListNode *&head)

    {

        ListNode *p=NULL;

        int i=0;

        int arr[10]= {9,8,5,4,4,3,3,3,2,1};

        for(i=0; i<3; i++)

        {

            if(head==NULL)

            {

                head=new ListNode(arr[i]);

                if(head==NULL)

                    return;

            }

            else

            {

                p=new ListNode(arr[i]);

                p->next=head;

                head=p;

            }

        }

    }

};



int main()

{

    Solution s;

    ListNode *L=NULL;

    s.createList(L);

    ListNode *head=L;

    while(head)

    {

        cout<<head->val<<" ";

        head=head->next;

    }

    cout<<endl;

    L=s.reverseBetween(L,2,3);

    while(L)

    {

        cout<<L->val<<" ";

        L=L->next;

    }

}