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206. Reverse Linked List

Description

Given the head of a singly linked list, reverse the list, and return the reversed list.

 

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

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<