328. Odd Even Linked List

Description

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
Note:

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

Problem URL

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Solution

讲一个链表变为奇数在前偶数在后的形式。

Using two pointers to solve this problem. For even node, it is after odd node, so in the while loop, even and even.next is used for stop criterion. Two pointers walk ahead. After the loop, let odd.next = evenhead.

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head != null){
            ListNode odd = head, even = head.next;
            ListNode evenHead = even;
            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;
    }
}

Time Complexity: O(n)
Space Complexity: O(1)

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