6

输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。

示例 1：
输入：head = [1,3,2]
输出：[2,3,1]

限制：
0 <= 链表长度 <= 10000

答案：头插法

1

function ListNode(val) {
    this.val = val;
    this.next = null;
}
const head = new ListNode()
const head1 = new ListNode()
const head2 = new ListNode()
const head3 = new ListNode()
Object.assign(head,{val:undefined,next:head1})
Object.assign(head1,{val:1,next:head2})
Object.assign(head2,{val:2,next:head3})
Object.assign(head3,{val:3,next:null})
var reversePrint = function(head) {
    const res = []
    function recur (ListNode){
        if(ListNode.next===null) {
            res.push(ListNode.val)
            return
        }
        recur(ListNode.next)
        if(ListNode.val)
            res.push(ListNode.val)
    }
    recur(head)
    return res
}
reversePrint(head)

2

var reversePrint = function (head) {
  if (head === null) return []
  const res = []
  while (head) {
    res.push(head.val)
    head = head.next
  }
  return res.reverse()
}

3

var reversePrint = function(head) {
     if (head === null) return []
    const res=[]
    while(head){
        res.unshift(head.val)
        head=head.next
    }
    return res
}

4

// 首先将链表反转
function reverseLink(head) {
    if (head === null || head.next === null) return head
    let p = head.next
    head.next = null
    let tmp = null
    while (p !== null) {
        tmp = p.next // tmp 指针前进（保存下一个指针信息）
        p.next = head // 指针反转
        head = p // head 指针前进
        p = tmp // p 指针前进
    }
    return head
}
//再顺序输出链表