Leetcode -- Odd Even Linked List

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Question:

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 ...

Analysis:

给出一个奇数偶数相交叉排列的链表，尝试将他们根据节点的奇偶性分开，而且要求空间复杂度为O(1)、时间复杂度为O(nodes).

考虑到时间、空间复杂度，我们只能遍历链表一次且使用有限的变量存储，因此我们可以首先将链表head分为奇数链表和偶数链表，然后每个链表都有一个头指针，最后只需要将奇数链表的尾节点链接到偶数链表的头结点即可。

Answer:

/**
 * 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 odd, even, evenhead;
         odd = head;
         even = head.next;
         evenhead = head.next;
         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;
    }
}

Leetcode -- Odd Even Linked List

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原文地址：http://www.cnblogs.com/little-YTMM/p/5184849.html