[LeetCode] 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.


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


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


這題是 024. Swap Nodes in Pairs 的進階題,不同於上一題是兩兩交換,這一題是以每 k 個節點為一組翻轉鍊結陣列。


現在有一陣列為 A->B->C->D->E ,現在我們要翻轉 BCD 三個節點,其步驟如下:
1. C->B
2. D->C
3. B->E
4. A->D

1. 先把 k 個節點翻轉,也就是步驟 1 與 2
2. 再將這 k 個節點與前後連起來,步驟 3 與 4。 


class Solution(object):
	def __init__(self,k=None):
		self.k = None
	def reverseKGroup(self, head, k):
		self.k = k
		if not head or self.k <= 1:
			return head

		dummy_head = ListNode(-1)
		dummy_head.next = head
		iteration = dummy_head

		while self.hasNextK(iteration):
			iteration = self.reverseNextK(iteration)

		return dummy_head.next

	def hasNextK(self, head):
		has = True
		tmp = head
		for i in range(self.k):
			if tmp.next is None :
				has = False
			tmp = tmp.next

		return has

	def reverseNextK(self, head): 
		pre = head
		current = head.next
		group_end = head.next

		for i in range(self.k):
			next_node = current.next
			current.next = pre
			pre, current  = current, next_node

		head.next = pre
		group_end.next = current

		return group_end

[LeetCode] 解題目錄 -> 這裡走



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