LeetCode 328. 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.

 

Example 1:

 

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]

Constraints:

• The number of nodes in the linked list is in the range [0, 104].
• -10^6 <= Node.val <= 10^6

 

연결 리스트 문제로, 홀수의 인덱스에 있는 노드와 짝수의 인덱스에 있는 노드를 합치는 문제이다.

 

static class Solution {
        public static ListNode oddEvenList(ListNode head) {
            if(head == null)
                return null;
            ListNode odd = head;
            ListNode evenHead = head.next;
            ListNode even = evenHead;
            while(even != null && even.next != null) {
                odd.next = odd.next.next;
                even.next = even.next.next;
                odd = odd.next;
                even = even.next;
            }
            odd.next = evenHead;
            return head;
        }
    }