328. Odd Even Linked List

Problem 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

Python Solution

class Solution:
    def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next:
            return head
            
        # Initialize pointers
        odd = head
        even = head.next
        even_head = even
        
        # Process nodes in pairs
        while even and even.next:
            # Connect odd nodes
            odd.next = even.next
            odd = odd.next
            
            # Connect even nodes
            even.next = odd.next
            even = even.next
        
        # Connect odd tail to even head
        odd.next = even_head
        
        return head

Java Solution

class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        // Initialize pointers
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = even;
        
        // Process nodes in pairs
        while (even != null && even.next != null) {
            // Connect odd nodes
            odd.next = even.next;
            odd = odd.next;
            
            // Connect even nodes
            even.next = odd.next;
            even = even.next;
        }
        
        // Connect odd tail to even head
        odd.next = evenHead;
        
        return head;
    }
}

C++ Solution

class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if (!head || !head->next) {
            return head;
        }
        
        // Initialize pointers
        ListNode* odd = head;
        ListNode* even = head->next;
        ListNode* evenHead = even;
        
        // Process nodes in pairs
        while (even && even->next) {
            // Connect odd nodes
            odd->next = even->next;
            odd = odd->next;
            
            // Connect even nodes
            even->next = odd->next;
            even = even->next;
        }
        
        // Connect odd tail to even head
        odd->next = evenHead;
        
        return head;
    }
};

JavaScript Solution

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var oddEvenList = function(head) {
    if (!head || !head.next) {
        return head;
    }
    
    // Initialize pointers
    let odd = head;
    let even = head.next;
    const evenHead = even;
    
    // Process nodes in pairs
    while (even && even.next) {
        // Connect odd nodes
        odd.next = even.next;
        odd = odd.next;
        
        // Connect even nodes
        even.next = odd.next;
        even = even.next;
    }
    
    // Connect odd tail to even head
    odd.next = evenHead;
    
    return head;
};

C# Solution

public class Solution {
    public ListNode OddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        // Initialize pointers
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = even;
        
        // Process nodes in pairs
        while (even != null && even.next != null) {
            // Connect odd nodes
            odd.next = even.next;
            odd = odd.next;
            
            // Connect even nodes
            even.next = odd.next;
            even = even.next;
        }
        
        // Connect odd tail to even head
        odd.next = evenHead;
        
        return head;
    }
}