-
Notifications
You must be signed in to change notification settings - Fork 0
Commit
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
- Loading branch information
1 parent
4e5446a
commit 3f0762a
Showing
1 changed file
with
44 additions
and
0 deletions.
There are no files selected for viewing
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,44 @@ | ||
Given the head of a singly linked list, reverse the list, and return the reversed list. | ||
|
||
|
||
|
||
Example 1: | ||
|
||
|
||
Input: head = [1,2,3,4,5] | ||
Output: [5,4,3,2,1] | ||
Example 2: | ||
|
||
|
||
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? | ||
|
||
best solution | ||
|
||
class Solution { | ||
public ListNode reverseList(ListNode head) { | ||
ListNode prev = null; | ||
ListNode curr = head; | ||
while (curr != null) { | ||
ListNode nextTemp = curr.next; | ||
curr.next = prev; | ||
prev = curr; | ||
curr = nextTemp; | ||
} | ||
return prev; | ||
} | ||
} | ||
|