给定一个链表,旋转链表,将链表每个节点向右移动 k 个位置,其中 k 是非负数。
示例 1:
输入: 1->2->3->4->5->NULL, k = 2
输出: 4->5->1->2->3->NULL
解释:
向右旋转 1 步: 5->1->2->3->4->NULL
向右旋转 2 步: 4->5->1->2->3->NULL
示例 2:
输入: 0->1->2->NULL, k = 4
输出: 2->0->1->NULL
解释:
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def rotateRight(self, head: ListNode, k: int) -> ListNode:
if head is None or head.next is None :
return head
cnt = 1
tmp = head
while tmp.next is not None:
tmp = tmp.next
cnt += 1
k = k%cnt
#print(cnt,k)
if k==0:
return head
tail1 = head
tail2 = head
for i in range(k):
tail2 = tail2.next
while tail2.next is not None:
tail1 = tail1.next
tail2 = tail2.next
#print(head.val,tail1.val,tail2.val)
tail2.next = head
head = tail1.next
tail1.next = None
return head