题目大意

将一个链表中每k个数进行翻转，末尾不足k个的数不做变化。

解题思路

参考：
https://shenjie1993.gitbooks.io/leetcode-python/025 Reverse Nodes in k-Group.html
A->B->C->D->E，现在我们要翻转BCD三个节点。进行以下几步：
1.C->B
2.D->C
3.B->E
4.A->D
5.返回及节点B

代码

class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head or k <= 1:
            return head
        dummy = ListNode(0)
        dummy.next = head
        current = dummy
        while current:
            current = self.reverseNextK(current, k)
        return dummy.next

    def reverseNextK(self, head, k):
        # Check if there are k nodes left
        current = head
        for i in range(k):
            if not current.next:
                return None
            current = current.next

        # The last node when the k nodes reversed
        node = head.next
        prev = head
        curr = head.next
        # Reverse k nodes
        for i in range(k):
            nextNode = curr.next
            curr.next = prev
            prev = curr
            curr = nextNode
        # Connect with head and tail
        node.next = curr
        head.next = prev
        return node

总结

太绕了