描述：
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->nullptr, m = 2 and n = 4,
return 1->4->3->2->5->nullptr.
Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.
代码：
class Solution {
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
ListNode dummy(-1);
dummy.next = head;
ListNode *prev = &dummy;
for (int i = 0; i < m-1; i++) {
prev = prev->next;
}
ListNode *head2 = prev;
prev = head2->next;
ListNode *cur = prev->next;
for (int i = m; i < n; i++) {
prev->next = cur->next;
cur->next = head2->next;
head2->next = cur; //头插法
cur = prev->next;
}
return dummy.next;
}
};
