链表"/>
LeetCode 61 旋转链表
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
解题思路:
闭合为环
力扣
【Leetcode】Python&C++:61. 旋转链表 (链表操作)【每日一题系列20210327】_哔哩哔哩_bilibili
Python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 k == 0 or not head or not head.next:return headcur = headn = 1while cur.next:cur = cur.nextn += 1add = n - k % nif add == n:return headcur.next = headwhile add:cur = cur.nextadd -= 1res = cur.nextcur.next = Nonereturn res
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LeetCode 61 旋转链表
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