Reverse K-group linked list
难度:Medium; 类型: linkedlist; 公司:字节研发,华为等
1. Links
2. 题目描述
将给出的链表中的节点每 k k 个一组翻转,返回翻转后的链表 如果链表中的节点数不是 k k 的倍数,将最后剩下的节点保持原样 你不能更改节点中的值,只能更改节点本身。 要求空间复杂度 O(1) O(1) 例如: 给定的链表是1→2→3→4→5 对于 k = 2 k=2, 你应该返回 2→1→4→3→5 对于 k = 3 k=3, 你应该返回3→2→1→4→5 示例1
输入
复制复制
{1,2,3,4,5},2
返回值
复制复制
{2,1,4,3,5}
3.思路: 分两步找每个group,以及把group翻转
找linkedlist的每个group的head和end
由于reverse后的linkedlist的head会变动,需要先fakehead使fakehead的next始终是list的head
之后用two pointer的方法,slowpointer指向group的head的前一个node,而fast 指向group的end在把fast不断向前移动时用counter记录它的位置来如果counter %k=0那么就是group的end
把group先cut出来进行reverse,然后返回新的group head 和end,再把它接上去原来的list
把 slow更新到fast
把每个group的head 和end的部分进行reverse
因为要对每一个node的next进行翻转,需要cur pointer 遍历每个node
需要prev保存cur的上一个node翻转,同时也要next保持cur的下一个node用于先前移动
new tail就是原来的list的head, new head就是最后的prev
4. Code
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
#
#
# @param head ListNode类
# @param k int整型
# @return ListNode类
#
class Solution:
def reverseKGroup(self , head , k ):
# write code here
#input: head of linkedlist, k of group size. output: new head of reverse linkedlist
#idea: 用two pointer method
# 物理意义: slow = head, fast = end of sub list
# 由于head会变动,不利于换行更新后的head,所以要加一个fakehead使得fakehead后面的next是新的head
# fast从左往右数计算当前的position% k来看 fast是不是一个size of k的group的end
# if so, reverse slow , fast 之间的sub link list
# then slow = fast, 更新下一个group的head
# base case: head none, return head no reverse
# Time: O(n) for iterate every node, O(k) for reverse group
# so O(kn)
# Space: O(1)
if not head or k <=1:
return head
fake_head = ListNode(None)
fake_head.next = head
slow = fake_head
fast = slow
cnt = 0
while fast:
if cnt!= 0 and cnt%k==0:
# cut sublist
next_head = fast.next
subls_head = slow.next
fast.next = None
new_head, new_tail = self.reverse(slow.next)
# connect
new_tail.next = next_head
slow.next = new_head
# update
fast = new_tail
slow = fast
# update position
fast = fast.next
cnt += 1
return fake_head.next
def reverse(self, node):
prev = None
cur = node
tail = node
while cur:
# backup and reverse
next_node = cur.next
cur.next = prev
# move forward
prev = cur
cur = next_node
new_head= prev
return new_head, tail
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