Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:

Input: 1->2->3->4->5->NULL, k = 2               Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
 

Example 2:Input: 0->1->2->NULL, k = 4 Output: 2->0->1->NULL

思路

　　對於鏈表類的題目最重要的就是指針的控制，因此對於這道題我的思路就是我們先找到倒數第K個節點前一個節點的位置，並先將鏈表尾部指針指向頭指針，然後將頭指針指向倒數第K個節點，最後對倒數K節點上一個指針賦值為None。時間復雜度為O(n)， 空間復雜度為O(1)

圖示步驟

解決代碼

 1 class Solution(object):
 2     def rotateRight(self, head, k):
 
 3         """
 4         :type head: ListNode
 5         :type k: int
 6         :rtype: ListNode
 7         """
 8         if not head or k < 0:      # 為空或者K小於0直接返回
 9             return head
10         length, tem = 0, head      
11         
12         while tem:                 # 求出鏈表的長度
 
13             tem, length = tem.next, length + 1
14             
15         k = k % length             # 防止K的長度大於鏈表的長度
16         fast, slow = head, head
17         while k > 0:              # 快指針先走K步
18             fast, k = fast.next, k-1
19         
20         while fast.next:          # 兩個指針同時動
21             slow, fast = slow.next, fast.next
22         fast.next = head            # 進行指針交換操作得到結果
23         head = slow.next
24         slow.next = None
25         return head
26         

【LeetCode每天一題】Rotate List(旋轉鏈表)