[LeetCode] Odd Even Linked List

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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

Note:
The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on…

解题思路

分别将奇数和偶数位置的结点连接起来，最后再把奇数和偶数链表连接起来。

实现代码

//Runtime: 1 ms
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */
public class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode oddTail = head;
        ListNode even = head.next;
        ListNode evenTail = even;
        int i = 1;
        ListNode cur = head.next.next;
        while (cur != null) {
            if (i++ % 2 == 1) {
                oddTail.next = cur;
                oddTail = oddTail.next;
            } else {
                evenTail.next = cur;
                evenTail = evenTail.next;
            }
            cur = cur.next;
        }
        evenTail.next = null;
        oddTail.next = even;

        return head;
    }
}