题目地址(61. 旋转链表)
题目描述
给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。
示例 1：
输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]
示例 2：
输入：head = [0,1,2], k = 4
输出：[2,0,1]
提示：
链表中节点的数目在范围 [0, 500] 内
-100 <= Node.val <= 100
0 <= k <= 2 * 109
思路
通过双指针，先把整个链表形成一个循环，然后计算移动到k步之后的链头，断开循环，形成新的链表
代码
- 语言支持：Python3
Python3 Code:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
if head == None:
return head
headPoint = head
length = 1
while head.next != None:
head = head.next
length += 1
endPoint = head
endPoint.next = headPoint #形成一个循环
moveLength = length - (k % length) - 1 #得出移动步数
print(moveLength)
while moveLength != 0:
headPoint = headPoint.next
moveLength -= 1
endPoint = headPoint
headPoint = headPoint.next
endPoint.next = None
return headPoint
复杂度分析
令 n 为数组长度。
- 时间复杂度：
- 空间复杂度：
