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
该题需要将链表转换为数组,且需要反向。由于目标是链表,无法第一时间得知长度,声明等长数组。
解题方案:
- 遍历
- 从头到尾遍链表,获取链表长度,声明等长数组;
- 再次遍历并放入数组当中,在数组中的放置顺序是从尾到头。
- 递归
- 记录深度,递归到链表尾部;
- 将深度化为数组长度,将回溯结果正序放入数组当中。
- 动态数组
- 遍历链表,将元素放入数组当中;
- 遍历结束,将数组倒置后返回(
reverse()
)。
# 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]
栈实现:
/**
* 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()];
int i = 0;
while (!stk.isEmpty()) {
ans[i++] = stk.pop();
}
return ans;
}
}
先计算链表长度 n,然后创建一个长度为 n 的结果数组。最后遍历链表,依次将节点值存放在数组上(从后往前):
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int[] reversePrint(ListNode head) {
if (head == null) {
return new int[]{};
}
int n = 0;
for (ListNode cur = head; cur != null; cur = cur.next, ++n);
int[] ans = new int[n];
for (ListNode cur = head; cur != null; cur = cur.next) {
ans[--n] = cur.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) {
let ans = [];
for (; !!head; head = head.next) {
ans.unshift(head.val);
}
return ans;
};
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reversePrint(head *ListNode) []int {
ans := []int{}
for head != nil {
ans = append([]int{head.Val}, ans...)
head = head.Next
}
return ans
}
递归实现:
/**
* 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;
}
};
栈实现:
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
stack<int> stk;
vector<int> ans;
ListNode *p = head;
while (p) {
stk.push(p->val);
p = p->next;
}
while (!stk.empty()) {
ans.push_back(stk.top());
stk.pop();
}
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;
}
动态数组:
// 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
}
}
遍历:
// 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
}
}