【LeetCode】61.旋转链表

1.题目

2. 分析

分割然后链接即可。

3. 代码

# 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 not head :
            return head
        length = 0
        s = head
        # 求出链表长度
        while(s):
            s = s.next
            length+=1        
        k %= length
        if k == 0:
            return head
        s = head
        split_head = head # 分割后的链表头部
        cnt = 0
        pre = None
        while(cnt < length - k):
            pre = split_head
            split_head = split_head.next
            cnt += 1
        pre.next = None # 这一步比较重要，否则后面会造成死循环
        tmp = split_head        
        while(split_head):
            pre = split_head
            split_head = split_head.next
        pre.next = s
        return