leetcode-25-Reverse-Nodes-in-k-Group

描述


Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list’s nodes, only nodes themselves may be changed.

Example 1:

img

1
2
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

img

1
2
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Example 3:

1
2
Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]

Example 4:

1
2
Input: head = [1], k = 1
Output: [1]

Constraints:

  • The number of nodes in the list is in the range sz.
  • 1 <= sz <= 5000
  • 0 <= Node.val <= 1000
  • 1 <= k <= sz

Follow-up: Can you solve the problem in O(1) extra memory space?

分析


链表翻转问题,每 k 个一组,翻转链表,适合练习。

解决方案1(Python)


1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverse(self, head, tail):
pre = tail.next
nowHead = head

while pre != tail:
nextNode = nowHead.next
nowHead.next = pre
pre = nowHead
nowHead = nextNode
return tail, head


def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:

dummy = ListNode(0, head)
pre = dummy

while head:
tail = pre
for i in range(k):
tail = tail.next
if not tail:
return dummy.next
nextNode = tail.next
head, tail = self.reverse(head, tail)
pre.next = head
tail.next = nextNode
pre = tail
head = tail.next
return dummy.next

解决方案2(Golang)


1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseKGroup(head *ListNode, k int) *ListNode {
dummy := &ListNode{Next: head}
pre := dummy

for head != nil {
tail := pre
for i := 0; i < k; i++ {
tail = tail.Next
if tail == nil {
return dummy.Next
}
}
next := tail.Next
head, tail = myReverse(head, tail)
pre.Next = head
tail.Next = next
pre = tail
head = tail.Next
}
return dummy.Next
}

func myReverse(head, tail *ListNode) (*ListNode, *ListNode) {
pre := tail.Next
now := head
for pre != tail {
next := now.Next
now.Next = pre
pre = now
now = next
}
return tail, head
}

相关问题


题目来源