lc25 k个一组反转单链表

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy;
        while (head != null) {
            ListNode tail = head;
            for (int i = 0; i < k - 1; i++) {
                tail = tail.next;
                if (tail == null) {
                    return dummy.next;
                }
            }
            ListNode nextHead = tail.next;
            ListNode[] reverse = reverse(head, tail);
            pre.next = reverse[0];
            reverse[1].next = nextHead;
            head = nextHead;
            pre = reverse[1];
        }
        return dummy.next;
    }

    public ListNode[] reverse(ListNode head, ListNode tail) {
        ListNode pre = null;
        // pre 可以赋值为tail.next；上面的reserve[1].next=nextHead的语义是一致的，把新尾部连接到原尾部的后继
        ListNode p = head;
        // 这里pre判断条件为链表的尾部，p的判断条件为链表尾部的后继，这次第一次做错在这里 mark一下
        while (pre != tail) {
            ListNode next = p.next;
            p.next = pre;
            pre = p;
            p = next;
        }
        return new ListNode[]{tail, head};
    }
}

链表经典题(说基础题确实不合适)，里面操作不复杂，思路细致一些即可