【LeetCode】旋转链表

目录

• 一、题目
• 二、解法
• 完整代码

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一、题目

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

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二、解法

先取余，化简一下k，然后找到倒数第k+1个节点，然后做断开重新连接的动作

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完整代码

# 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
        dummy = cur = ListNode()
        dummy.next = head
        tot = 0
        while cur.next:
            cur = cur.next
            tot += 1
        last = cur
        cur = dummy
        k %= tot
        tmp = cur
        for _ in range(k + 1):
            tmp = tmp.next
        kk = cur
        while tmp:
            kk = kk.next
            tmp = tmp.next
        newHead = kk.next
        if not newHead:
            return dummy.next
        last.next = head
        kk.next = None
        return newHead

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