【leetcode】Reverse Linked List II

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.
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
思路：
找到n和m两个点，然后再一对对地交换其val.

class Solution {public:    ListNode* reverseBetween(ListNode* head, int m, int n) {        ListNode *p,*q,*re,*ne;        p=head;        for(int i=1;i<m;i++) //注意从1开始，因为之前已经指向了head，少了一个点。        {            p=p->next;        }        for(int i=m;i<n;i++)        {            q=p;            for(int j=i;j<n;j++)                q=q->next;            swap(p->val,q->val);            n--;            p=p->next;        }        return head;    }};