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
方法一：顺序遍历 + 反转
我们可以顺序遍历链表，将每个节点的值存入数组中，然后将数组反转。
时间复杂度
方法二：递归
我们可以使用递归的方式，先递归得到 head 之后的节点反过来的值列表，然后将 head 的值加到列表的末尾。
时间复杂度
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
ans = []
while head:
ans.append(head.val)
head = head.next
return ans[::-1]# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reversePrint(self, head: ListNode) -> List[int]:
if head is None:
return []
ans = self.reversePrint(head.next)
ans.append(head.val)
return ans/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int[] reversePrint(ListNode head) {
Deque<Integer> stk = new ArrayDeque<>();
for (; head != null; head = head.next) {
stk.push(head.val);
}
int[] ans = new int[stk.size()];
for (int i = 0; !stk.isEmpty(); ++i) {
ans[i] = stk.pop();
}
return ans;
}
}/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int[] reversePrint(ListNode head) {
int n = 0;
ListNode cur = head;
for (; cur != null; cur = cur.next) {
++n;
}
int[] ans = new int[n];
cur = head;
for (; cur != null; cur = cur.next) {
ans[--n] = cur.val;
}
return ans;
}
}/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
if (!head) return {};
vector<int> ans = reversePrint(head->next);
ans.push_back(head->val);
return ans;
}
};/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
vector<int> ans;
for (; head; head = head->next) {
ans.push_back(head->val);
}
reverse(ans.begin(), ans.end());
return ans;
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reversePrint(head *ListNode) (ans []int) {
for ; head != nil; head = head.Next {
ans = append(ans, head.Val)
}
for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
ans[i], ans[j] = ans[j], ans[i]
}
return
}/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reversePrint(head *ListNode) (ans []int) {
if head == nil {
return
}
ans = reversePrint(head.Next)
ans = append(ans, head.Val)
return
}/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {number[]}
*/
var reversePrint = function (head) {
let ans = [];
for (; !!head; head = head.next) {
ans.unshift(head.val);
}
return ans;
};/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {number[]}
*/
var reversePrint = function (head) {
if (!head) {
return [];
}
const ans = reversePrint(head.next);
ans.push(head.val);
return ans;
};/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function reversePrint(head: ListNode | null): number[] {
let ans: number[] = [];
for (; !!head; head = head.next) {
ans.unshift(head.val);
}
return ans;
}// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
impl Solution {
pub fn reverse_print(head: Option<Box<ListNode>>) -> Vec<i32> {
let mut arr: Vec<i32> = vec![];
let mut cur = head;
while let Some(node) = cur {
arr.push(node.val);
cur = node.next;
}
arr.reverse();
arr
}
}// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
impl Solution {
pub fn reverse_print(head: Option<Box<ListNode>>) -> Vec<i32> {
let mut cur = &head;
let mut n = 0;
while let Some(node) = cur {
cur = &node.next;
n += 1;
}
let mut arr = vec![0; n];
let mut cur = head;
while let Some(node) = cur {
n -= 1;
arr[n] = node.val;
cur = node.next;
}
arr
}
}/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public int[] ReversePrint(ListNode head) {
List<int> ans = new List<int>();
while (head != null) {
ans.Add(head.val);
head = head.next;
}
ans.Reverse();
return ans.ToArray();
}
}