Reverse Linked List II

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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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Have you met this question in a real interview?

思路：对于一个链表： head->...->headstart->headtail(第m位)->第m+1位->...->pre(第n位)->cur（第n+1位）->.....->NULL

我们只需要略过链表前m-1个结点，将[m,n]中的结点反转，headtail指向cur完成没有反转结点的链接。

值得注意的是，headstart->next=pre(如果headstart不为空的话)

反转后的链表为：

(1)head->...->headstart->pre>...->headtail->cur（第n+1位）->.....->NULL(headstart!=NULL)

(2)pre>...->headtail->cur（第n+1位）->.....->NULL(headstart==NULL)

/** * 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||head->next==NULL)            return head;        if(m==n)            return head;        int num=1;        ListNode * pre=NULL;        ListNode *  cur=NULL;        ListNode *  post=NULL;        ListNode * headstart=NULL;        ListNode * headend=head;        ListNode * tailstart=NULL;        while(num<m)        {            headstart=headend;            headend=headend->next;            num++;        }                pre=headend;                tailstart=pre;        cur=pre->next;        pre->next=NULL;        while(num<n)        {                 post=cur->next;                cur->next=pre;                            pre=cur;                cur=post;                num++;        }        tailstart->next=cur;        if(headstart!=NULL)            headstart->next=pre;        else            head=pre;        return head;                    }};

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