25. K 个一组翻转链表

https://leetcode-cn.com/problems/reverse-nodes-in-k-group/

给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。

k 是一个正整数，它的值小于或等于链表的长度。

如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。

示例：

给你这个链表：1->2->3->4->5

当 k = 2 时，应当返回: 2->1->4->3->5

当 k = 3 时，应当返回: 3->2->1->4->5

说明：

你的算法只能使用常数的额外空间。 你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。

备注

• 时间复杂度：O(n)

Same as the iterative solution. It is because that in each function call, we need to first check if there are k nodes left starting from input head, and if there is, we have to reverse them. So two linear scans are still necessary.

• 空间复杂度：O(n/k)

With recursion, we need to take the cost of call stacks into consideration. For every k nodes, we are supposed to recursively call the function for once. So the total number of recursive calls is n / k. Within each function, we are still using a couple of pointers only in this algorithm, so they are constant cost. Thus, the total space complexity is O(n / k).

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head or k <= 0:
           return
        node_a = node_b = head
        for _ in range(k):
            if not node_b:
                return head
            node_b = node_b.next

        # 上面遍历完成后，node_b 是下一组的头结点，统一也是翻转部分连接的退出条件，等价于正常翻转数组的null
        new_head = self.reverse(node_a, node_b)
        node_a.next = self.reverseKGroup(node_b, k)
        return new_head

    def reverse(self, node_a, node_b):
        pre = None
        cur = node_a
        while cur is not node_b:
            tmp = cur.next
            cur.next = pre
            pre = cur
            cur = tmp
        return pre