Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given1->2->3->4->5->NULL, m = 2 and n = 4,
return1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
要定义一个链表头结点的指针，因为有可能反转的第一个节点是head；
不能采用从发转的第一个元素开始向后插入的方式，否则当要发转的元素个数为奇数时，{1,2,3,4,5}，2,4；输出结果会是{1,4,2,3,5}
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if(head==null)
return null;
int diff=n-m;
ListNode headPre=new ListNode(0);
headPre.next=head;
ListNode slow=head;
ListNode slowPre=headPre;
for(int i=1;i
