Easy
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:
[0, 5000]
.-5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
import com_github_leetcode.ListNode
/*
* Definition for singly-linked list.
* class ListNode(_x: Int = 0, _next: ListNode = null) {
* var next: ListNode = _next
* var x: Int = _x
* }
*/
object Solution {
def reverseList(head: ListNode): ListNode = {
var prev: ListNode = null
var curr = head
while (curr != null) {
val next = curr.next
curr.next = prev
prev = curr
curr = next
}
prev
}
}