leetcode之92. Reverse Linked List II（C++读错题版本，交换一个链表中指定的两个位置上的元素）

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.
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真的是没有读懂题啊，看例题真的是以为两个位置交换啊，英语不好好坑爹啊，程序写的也是乱，凑合看吧，意思还是这个意思，嘻嘻~~~~~~~~
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/** * 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 || !head->next ||m==n) return head;    ListNode *p=head;    ListNode *q=head;    ListNode *pm;    ListNode *pn;    ListNode *pmp;    ListNode *pnp;      if(!head->next->next)        {            ListNode *second=head->next;            second->next=head;            head->next=nullptr;            return second;        }        else          if(1==m && 2==n)        {            ListNode *seconds=head->next;            ListNode *third=head->next->next;            seconds->next=head;            head->next=third;            return seconds;        }        else    if(1==m)    {        int j=0;        ListNode *pnp1;        ListNode *pm1=head;        ListNode *pn1;        ListNode *p1=head;        while(p1)        {            if(j==n-2)            {                pnp1=p1;                pn1=p1->next;                break;            }            else ;            p1=p1->next;            j++;        }        ListNode *temp1=pm1->next;        pm1->next=pn1->next;        pn1->next=temp1;        if(pnp1==temp1)        {            pnp1->next=pm1;            return pn1;        }        else        {            pnp1->next=pm1;            return pn1;        }    }    else    {        int i=0;        while(p)        {            if(i==m-2)            {                pmp=p;                pm=p->next;            }            else if(i==n-2)            {                pnp=p;                pn=p->next;                break;            }            else ;            p=p->next;            i++;        }//i 就是链表的长度        pmp->next=pn;        ListNode *temp=pm->next;        pm->next=pn->next;        if(temp==pn)        {            pn->next=pm;            return head;        }        else        {            pn->next=temp;            pnp->next=pm;        }        return head;    }}};

下一篇更正确版本，改不对宝宝不睡了！