Odd Even Linked List

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.

For example:

• Input: head = [1,2,3,4,5]

• Output: [1,3,5,2,4]

• Input: head = [2,1,3,5,6,4,7]

• Output: [2,3,6,7,1,5,4]

# 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 not head:
            return head

        odd = head
        even = head.next
        even_head = even

        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next

        odd.next = even_head
        return head

Explanation:

This Python code rearranges a singly linked list so that all nodes with odd indices appear before nodes with even indices. The first node is considered odd, the second is even, and so on. The relative order within the odd and even groups is preserved.

1. Initialization: The code starts by handling the base case where the list is empty. It then initializes pointers odd to the head of the list, even to the second node, and even_head to the second node to keep track of the beginning of the even list.

2. Iteration: The while loop continues as long as there are more even nodes to process (i.e., even is not None and even.next is not None). Inside the loop:

• The next odd node is connected to the node after the current even node: odd.next = even.next
• The odd pointer moves to the next odd node: odd = odd.next
• The next even node is connected to the node after the current odd node: even.next = odd.next
• The even pointer moves to the next even node: even = even.next

1. Concatenation: After the loop, the last odd node is connected to the head of the even list: odd.next = even_head

2. Return: The function returns the modified head of the list.

Brute Force Solution:

A brute-force approach might involve creating two separate lists, one for odd-indexed nodes and one for even-indexed nodes, and then concatenating them. This would require iterating through the list twice and would not satisfy the O(1) space complexity requirement.

Time Complexity:

The time complexity of this solution is O(n), where n is the number of nodes in the linked list. This is because the code iterates through the list once.

Space Complexity:

The space complexity is O(1) because the code uses a constant amount of extra space, regardless of the size of the input list. It only uses a few pointers to keep track of the nodes.

Edge Cases:

• Empty List: If the input list is empty, the function should return None.
• List with One Node: If the list has only one node, it is already sorted, so the function should return the head.
• List with Two Nodes: If the list has two nodes, the function should swap the nodes and return the head.
• List with an Even Number of Nodes: The function should correctly handle lists with an even number of nodes.
• List with an Odd Number of Nodes: The function should correctly handle lists with an odd number of nodes.