Description#
Given the head
of a singly linked list, reverse the list, and return the reversed list.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0206.Reverse%20Linked%20List/images/rev1ex1.jpg)
Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0206.Reverse%20Linked%20List/images/rev1ex2.jpg)
Input: head = [1,2]
Output: [2,1]
Example 3:
Input: head = []
Output: []
Constraints:
- The number of nodes in the list is the range
[0, 5000]
. -5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
Solutions#
Solution 1#
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| # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
dummy = ListNode()
curr = head
while curr:
next = curr.next
curr.next = dummy.next
dummy.next = curr
curr = next
return dummy.next
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| /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode dummy = new ListNode();
ListNode curr = head;
while (curr != null) {
ListNode next = curr.next;
curr.next = dummy.next;
dummy.next = curr;
curr = next;
}
return dummy.next;
}
}
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| /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* dummy = new ListNode();
ListNode* curr = head;
while (curr) {
ListNode* next = curr->next;
curr->next = dummy->next;
dummy->next = curr;
curr = next;
}
return dummy->next;
}
};
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| /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
dummy := &ListNode{}
curr := head
for curr != nil {
next := curr.Next
curr.Next = dummy.Next
dummy.Next = curr
curr = next
}
return dummy.Next
}
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| /**
* 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 reverseList(head: ListNode | null): ListNode | null {
if (head == null) {
return head;
}
let pre = null;
let cur = head;
while (cur != null) {
const next = cur.next;
cur.next = pre;
[pre, cur] = [cur, next];
}
return pre;
}
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| // 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_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
let mut head = head;
let mut pre = None;
while let Some(mut node) = head {
head = node.next.take();
node.next = pre.take();
pre = Some(node);
}
pre
}
}
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| /**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var reverseList = function (head) {
let dummy = new ListNode();
let curr = head;
while (curr) {
let next = curr.next;
curr.next = dummy.next;
dummy.next = curr;
curr = next;
}
return dummy.next;
};
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| /**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public ListNode ReverseList(ListNode head) {
ListNode pre = null;
for (ListNode p = head; p != null;)
{
ListNode t = p.next;
p.next = pre;
pre = p;
p = t;
}
return pre;
}
}
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Solution 2#
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| # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
ans = self.reverseList(head.next)
head.next.next = head
head.next = None
return ans
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| /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode ans = reverseList(head.next);
head.next.next = head;
head.next = null;
return ans;
}
}
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| /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (!head || !head->next) return head;
ListNode* ans = reverseList(head->next);
head->next->next = head;
head->next = nullptr;
return ans;
}
};
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| /**
* 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
}
ans := reverseList(head.Next)
head.Next.Next = head
head.Next = nil
return ans
}
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| /**
* 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)
* }
* }
*/
const rev = (pre: ListNode | null, cur: ListNode | null): ListNode | null => {
if (cur == null) {
return pre;
}
const next = cur.next;
cur.next = pre;
return rev(cur, next);
};
function reverseList(head: ListNode | null): ListNode | null {
if (head == null) {
return head;
}
const next = head.next;
head.next = null;
return rev(head, next);
}
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| // 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 {
fn rev(pre: Option<Box<ListNode>>, cur: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
match cur {
None => pre,
Some(mut node) => {
let next = node.next;
node.next = pre;
Self::rev(Some(node), next)
}
}
}
pub fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
Self::rev(None, head)
}
}
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