【LeetCode】0025. 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:

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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.


Related Topics: Linked List



解題邏輯與實作

這題是 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 個節點為一組翻轉鏈結串列。

不過這題在實做時,卡了我老久,主要是我試圖在一個迴圈把所整個翻轉過程,連同頭尾的指標一併搞定,所以寫起來超級的不順手。後來在這一篇中看到它的整理,這才總算讓我的思考清晰了許多:

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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

整的步驟其實分成兩個部份:
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。 
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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
            break
         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



其他連結

  1. 【LeetCode】0000. 解題目錄