Given the head of a singly linked list, reverse the list, and return the reversed list.
Example 1:
1 -> 2 -> 3 -> 4 -> 5
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v
5 -> 4 -> 3 -> 2 -> 1
Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
Example 2:
1 -> 2
2 -> 1
Input: head = [1,2]
Output: [2,1]
Example 3:
Input: head = []
Output: []
Constraints:
- The number of nodes in the list is the range
[0, 5000].
-5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
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# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
''' Recursive '''
class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head or not head.next:
return head
next_node = self.reverseList(head.next)
head.next.next = head
head.next = None
return next_node
''' Iterative '''
class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev = None
while head:
new_head = head.next
head.next = prev
prev = head
head = new_head
return prev
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/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
/* Recursive */
func reverseList(head *ListNode) *ListNode {
if head == nil || head.Next == nil{
return head
}
newHead := reverseList(head.Next)
head.Next.Next = head
head.Next = nil
return newHead
}
/* Iterative */
func reverseList(head *ListNode) *ListNode {
var prev *ListNode
nextHead := head
for head != nil {
nextHead = head.Next
head.Next = prev
prev = head
head = nextHead
}
return prev
}
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