328. Odd Even Linked List

Description of Problem

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, 10^4].
-10^6 <= Node.val <= 10^6

Solution

Tags: LinkedList

Explanation

Please look at LeetCoder25's Solution

Code (C++)

/**
 * 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 == nullptr || head->next == nullptr){
            return head;
        }

        ListNode * odd = head;
        ListNode * evenHead = head->next;
        ListNode * even = evenHead;

        while(odd->next != nullptr && even->next != nullptr){
            odd->next = even->next;
            even->next = odd->next->next;
            odd = odd->next;
            even = even->next;
        }
        odd->next = evenHead;
        return head;
    }
};

Complexity

n is the length of the linkedlist

Time Complexity

\( T(n) = O(n) \)

Auxiliary Space

\( S(n) = O(1) \)

Reference

LeetCoder25's Solution

Leetbook