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【LeetCode】61. 旋转链表
1 问题
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]
2 答案
自己写的不对
class Solution:def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:l = 0dunny = ListNode(0)dunny.next = headwhile head:head = head.nextl += 1if k % l == 0:return headfor _ in range(k % l):while head:if head.next == None:head.next = dunny.nextbreakhead = head.nextreturn dunny.next
官方解,使用快慢指针
# 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 or not head.next: return headl = 0 # 长度cur = head # 要复制一份,不然求head的长度时,head就被遍历没了while cur:cur = cur.nextl += 1k %= lif k == 0: return headfast, slow = head, headwhile k: # 先让fast走k步fast = fast.nextk -= 1while fast.next: # 让fast走到倒数第一个节点,不让他走完fast = fast.nextslow = slow.nextnewHead = slow.next # slow的下一个节点是newHead的第一个节点slow.next = None # 对slow后进行打断fast.next = head # fast连接head的第一个节点return newHead
步骤:
- 求链表长度;
- 找出倒数第 k + 1 k+1 k+1 个节点;
- 链表重整:将链表的倒数第 k + 1 k+1 k+1 个节点和倒数第 k k k 个节点断开,并把后半部分拼接到原链表的头部。新链表的起始节点就是原链表的倒数第 k k k 个节点
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【LeetCode】61. 旋转链表
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