025 Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example：

Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5

Note：

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

解释下题目：

只能用交换节点的方法来逆转链表

1. 老老实实换呗

实际耗时：8ms

public ListNode reverseKGroup(ListNode head, int k) {
        ListNode listNode = new ListNode(0);
        ListNode start = listNode;
        listNode.next = head;
        while (true) {
            ListNode p = start;
            ListNode q;
            ListNode cur = p;
            start = p.next;
            for (int i = 0; i < k && cur != null; i++) {
                cur = cur.next;
            }
            if (cur == null) {
                break;
            }
            for (int i = 0; i < k - 1; i++) {
                q = p.next;
                p.next = q.next;
                q.next = cur.next;
                cur.next = q;
            }
        }
        return listNode.next;
    }

  思路见图：

时间复杂度O(n)

空间复杂度O(1)