K个一组翻转链表

思路: 先写反转 [head, tail) 的函数, 随后主函数先反转前 K 个链表, 再将该链表的头节点 (反转之后变成了尾节点) 的 next 设置为递归的返回值 (新的头节点)

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode tail = head;
        for (int i = 0; i < k; i++) {
            if (tail == null) return head;
            tail = tail.next;
        }
        // [head, tail)
        ListNode newHead = reverse(head, tail);
        head.next = reverseKGroup(tail, k);
        return newHead;
    }
 
    private ListNode reverse(ListNode head, ListNode tail) {
        ListNode pre = null, curr = head, nxt = head;
        while (curr != tail) {
            nxt = nxt.next;
            curr.next = pre;
            pre = curr;
            curr = nxt;
        }
        return pre;
    }
}