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328. Odd Even Linked List

Description

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

 

Example 1:

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

Example 2:

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

 

Constraints:

  • The number of nodes in the linked list is in the range [0, 104].
  • -106 <= Node.val <= 106

Solutions

  • /**
 * 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 oddEvenList(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode a = head;
        ListNode b = head.next, c = b;
        while (b != null && b.next != null) {
            a.next = b.next;
            a = a.next;
            b.next = a.next;
            b = b.next;
        }
        a.next = c;
        return head;
    }
}

  • /**
 * 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* oddEvenList(ListNode* head) {
        if (!head) {
            return nullptr;
        }
        ListNode* a = head;
        ListNode *b = head->next, *c = b;
        while (b && b->next) {
            a->next = b->next;
            a = a->next;
            b->next = a->next;
            b = b->next;
        }
        a->next = c;
        return head;
    }
};

  • # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None:
            return None
        a = head
        b = c = head.next
        while b and b.next:
            a.next = b.next
            a = a.next
            b.next = a.next
            b = b.next
        a.next = c
        return head


  • /**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func oddEvenList(head *ListNode) *ListNode {
	if head == nil {
		return nil
	}
	a := head
	b, c := head.Next, head.Next
	for b != nil && b.Next != nil {
		a.Next = b.Next
		a = a.Next
		b.Next = a.Next
		b = b.Next
	}
	a.Next = c
	return head
}

  • /**
 * 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 oddEvenList(head: ListNode | null): ListNode | null {
    if (!head) {
        return null;
    }
    let [a, b, c] = [head, head.next, head.next];
    while (b && b.next) {
        a.next = b.next;
        a = a.next;
        b.next = a.next;
        b = b.next;
    }
    a.next = c;
    return head;
}

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