328. Odd Even Linked List

题目分析

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 ...

注意题目说的是结点而不是结点中的值。

代码

解法一

/**
 * 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 || head.next == null) {
            return head;
        }
        ListNode t1 = head;
        ListNode dummy1 = new ListNode(0);
        ListNode dummy2 = new ListNode(0);
        ListNode temp1 = dummy1;
        ListNode temp2 = dummy2;
        // 判断是不是偶数
        boolean flag = true;
        while(t1 != null) {
            if(flag) {
                temp2.next = t1;
                temp2 = temp2.next;
                t1 = t1.next;
                temp2.next = null;
            } else {
                temp1.next = t1;
                temp1 = temp1.next;
                t1 = t1.next;
                temp1.next = null;
            }
            flag = !flag;
        }
        temp2.next = dummy1.next;
        return dummy2.next;
    }
}

解法二

/**
 * 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 || head.next == null) {
            return head;
        }
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = even;
        while(even != null && even.next != null) {
            odd.next = even.next;
            odd = odd.next;
            even.next = odd.next;
            even = even.next;
        }
        odd.next = evenHead;
        return head;
    }
}