25. Reverse Nodes in k-Group

Recursion

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

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

        ListNode dummyHead = new ListNode(-1, head);
        ListNode pre = dummyHead;

        while (pre.next != null) {
            ListNode currHead = pre.next;
            ListNode curr = currHead;
            for (int i = 0; i < k - 1 && curr != null; i++) {
                curr = curr.next;
            }
            if (curr == null) {
                // 不足 k 个，直接返回
                break;
            }

            ListNode nextHead = curr.next;
            curr.next = null;

            ListNode reversedHead = reverse(currHead);
            // d 3 2 1 4 5
            //   r   ch

            pre.next = reversedHead;
            currHead.next = nextHead;

            pre = currHead;
        }

        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;
    }
}

Reference

25. Reverse Nodes in k-Group