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3263. Convert Doubly Linked List to Array I 🔒

Description

You are given the head of a doubly linked list, which contains nodes that have a next pointer and a previous pointer.

Return an integer array which contains the elements of the linked list in order.

 

Example 1:

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

Output: [1,2,3,4,3,2,1]

Example 2:

Input: head = [2,2,2,2,2]

Output: [2,2,2,2,2]

Example 3:

Input: head = [3,2,3,2,3,2]

Output: [3,2,3,2,3,2]

 

Constraints:

  • The number of nodes in the given list is in the range [1, 50].
  • 1 <= Node.val <= 50

Solutions

Solution 1: Direct Traversal

We can directly traverse the linked list, adding the values of the nodes to the answer array $\textit{ans}$ one by one.

After the traversal is complete, return the answer array $\textit{ans}$.

The time complexity is $O(n)$, where $n$ is the length of the linked list. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

  • /*
// Definition for a Node.
class Node {
    public int val;
    public Node prev;
    public Node next;
};
*/

class Solution {
    public int[] toArray(Node head) {
        List<Integer> ans = new ArrayList<>();
        for (; head != null; head = head.next) {
            ans.add(head.val);
        }
        return ans.stream().mapToInt(i -> i).toArray();
    }
}


  • /**
 * Definition for doubly-linked list.
 * class Node {
 *     int val;
 *     Node* prev;
 *     Node* next;
 *     Node() : val(0), next(nullptr), prev(nullptr) {}
 *     Node(int x) : val(x), next(nullptr), prev(nullptr) {}
 *     Node(int x, Node *prev, Node *next) : val(x), next(next), prev(prev) {}
 * };
 */
class Solution {
public:
    vector<int> toArray(Node* head) {
        vector<int> ans;
        for (; head; head = head->next) {
            ans.push_back(head->val);
        }
        return ans;
    }
};


  • """
# Definition for a Node.
class Node:
    def __init__(self, val, prev=None, next=None):
        self.val = val
        self.prev = prev
        self.next = next
"""


class Solution:
    def toArray(self, root: "Optional[Node]") -> List[int]:
        ans = []
        while root:
            ans.append(root.val)
            root = root.next
        return ans


  • /**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Next *Node
 *     Prev *Node
 * }
 */

func toArray(head *Node) (ans []int) {
	for ; head != nil; head = head.Next {
		ans = append(ans, head.Val)
	}
	return
}


  • /**
 * Definition for _Node.
 * class _Node {
 *     val: number
 *     prev: _Node | null
 *     next: _Node | null
 *
 *     constructor(val?: number, prev? : _Node, next? : _Node) {
 *         this.val = (val===undefined ? 0 : val);
 *         this.prev = (prev===undefined ? null : prev);
 *         this.next = (next===undefined ? null : next);
 *     }
 * }
 */

function toArray(head: _Node | null): number[] {
    const ans: number[] = [];
    for (; head; head = head.next) {
        ans.push(head.val);
    }
    return ans;
}

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