Discuss：www.cnblogs.com/grandyang/p…
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?
解法一：
迭代求解。
/**
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/
class Solution {
fun reverseList(head: ListNode?): ListNode? {
if (head?.next == null) {
return head
}
var cur: ListNode? = head
var pre: ListNode? = null
while (cur != null) {
val tempNode = cur.next
cur.next = pre
pre = cur
cur = tempNode
}
return pre
}
}
解法二：
递归求解。
/**
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/
class Solution {
fun reverseList(head: ListNode?): ListNode? {
if (head?.next == null) {
return head
}
val listNode = reverseList(head.next)
head.next?.next = head
head.next = null
return listNode
}
}
