给你单链表的头节点
head,请你反转链表,并返回反转后的链表。
输入: head = [1,2,3,4,5]
输出: [5,4,3,2,1]
第一种解法
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
var prev *ListNode
curr := head
for curr != nil {
next := curr.Next
curr.Next = prev
prev = curr
curr = next
}
return prev
}
第二种解法
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
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
}