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?
非递归
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head){
if(head == null || head.next ==null){
return head;
}
ListNode pre = null;
ListNode now = head;
while(now!=null){
ListNode next = now.next;
now.next =pre;
pre = now;
now = next;
}
return pre;
}
}递归
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head){
if(head == null || head.next == null){
return head;
}
ListNode last = reverseList(head.next);
head.next.next = head;
head.next = null;
return last;
}
}