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.
题目大意就是给一个单链表，给定一个起点和一个终点，反转起点和终点之间的链表，要求：原地反转，一次遍历。
解题思路：因为已经限制了1 ≤ m ≤ n ≤ length of list，设定头节点指向第一个元素，工作指针先走m步，采用头插法来重新构建m到n之间的数据，最后把m位置上的next指针指向原来的n+1的位置。
这次代码写的有点纠结，Java操作这些不如c++熟练。
public ListNode reverseBetween(ListNode head, int m, int n) {
if (head == null || m == n) {
return head;
}
int count = n - m;
ListNode p = new ListNode(0);
ListNode q;
ListNode headp = new ListNode(0);
ListNode res = headp;
p.next = head;
headp.next = head;
while (--m > 0) {
headp = headp.next;
}
p = headp.next;
q = p.next;
ListNode last = p;
headp.next = null;
while (count-- >= 0) {
p.next = headp.next;
headp.next = p;
if (q != null) {
p = q;
q = q.next;
} else {
p = null;
}
}
last.next = p;
/*while(res!=null){
System.out.println(res.val);
res=res.next;
}*/
return res.next;
}