25. Reverse Nodes in k-Group

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Recursion

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class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        // 1 2 3 4 5, when k == 3

        ListNode curr = head;
        for (int i = 0; i < k - 1 && curr != null; i++) {
            curr = curr.next;
        }
        if (curr == null) {
            // 不足 k 个无需翻转，curr 停留在第 k 个节点上
            return head;
        }

        ListNode nextHead = curr.next;
        curr.next = null;
        ListNode newHead = reverse(head);
        head.next = reverseKGroup(nextHead, k);
        return newHead;
    }

    private ListNode reverse(ListNode head) {
        ListNode pre = null, curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = pre;
            pre = curr;
            curr = next;
        }

        return pre;
    }
}

Iterate

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class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummyHead = new ListNode(-1, head);
        ListNode pre = dummyHead;

        while (pre.next != null) {
            // p -> 1 -> 2 -> 3
            ListNode subTail = pre;
            for (int i = 0; i < k && subTail != null; i++) {
                subTail = subTail.next;
            }
            if (subTail == null) {
                break;
            }

            ListNode nextHead = subTail.next;
            subTail.next = null; // 不要忘记切断链表

            ListNode newSubHead = subTail;
            ListNode newSubTail = pre.next;
            reverse(pre.next);
            pre.next = newSubHead;

            newSubTail.next = nextHead;
            pre = newSubTail;
        }

        return dummyHead.next;
    }

    private ListNode reverse(ListNode head) {
        ListNode pre = null, curr = head;

        while (curr != null) {
            ListNode next = curr.next;
            curr.next = pre;
            pre = curr;
            curr = next;
        }

        return pre;
    }
}

References

25. Reverse Nodes in k-Group