206. 反转链表

206. 反转链表

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。

示例 1:

输入:head = [1,2,3,4,5]
输出:[5,4,3,2,1]

示例 2:

输入:head = [1,2]
输出:[2,1]

示例 3:

输入:head = []
输出:[]

提示:

  • 链表中节点的数目范围是 [0, 5000]
  • -5000 <= Node.val <= 5000

解法一:

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
    cur := head
    var pre *ListNode
    for cur != nil {
        temp := cur.Next
        cur.Next = pre
        pre = cur 
        cur = temp
    }
    return pre
}

 解法二:(递归)

func reverseList(head *ListNode) *ListNode {
    return help(nil, head)
}

func help(pre, head *ListNode)*ListNode{
    if head == nil {
        return pre
    }
    next := head.Next
    head.Next = pre
    return help(head, next)
}

 

posted on 2022-10-14 17:35  HHHuskie  阅读(12)  评论(0编辑  收藏  举报

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