reverse node 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.

# 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 head is None or k < 2:
            return head

        next_head = head
        for i in range(k - 1):
            next_head = next_head.next
            if next_head is None:
                return head
        ret = next_head

        current = head
        while next_head:
            tail = current
            prev = None
            for i in range(k):
                if next_head:
                    next_head = next_head.next
                _next = current.next
                current.next = prev
                prev = current
                current = _next
            tail.next = next_head or current

        return ret

再來一個(gè)直觀的：

先看Leetcode 206的reverse linkedlist的問題：

class Solution:
    def reverseList(self, head):
        if not head or not head.next:
            return head

        prev, cur, nxt = None, head, head
        while cur:
            nxt = cur.next
            cur.next = prev
            prev = cur
            cur = nxt
        return prev    

然后，問題可以轉(zhuǎn)化一下：

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

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        count, node = 0, head
        while node and count < k:
            node = node.next
            count += 1
        if count < k: return head
        new_head, prev = self.reverse(head, count)
        head.next = self.reverseKGroup(new_head, k)
        return prev

    def reverse(self, head, count):
        prev, cur, nxt = None, head, head
        while count > 0:
            nxt = cur.next
            cur.next = prev
            prev = cur
            cur = nxt
            count -= 1
        return (cur, prev)