LeetCode Odd Even Linked List

题意：给出一个单链表，将位于奇数位和偶数位的分组，重新组成链表，要求空间复杂度为O(1)

思路：用四个结点，其中两个表示奇数链表的头和尾，另外两个表示偶数链表的头和尾，注意，偶数链表的尾可能为空或者不为空，在不为空时，需要将下一结点置为空

代码如下：

class Solution
{public ListNode oddEvenList(ListNode head){if (null == head) return head;ListNode oddhead = null;ListNode oddtail = null;ListNode evenhead = null;ListNode eventail = null;ListNode p = head;int cnt = 1;while (p != null){if (1 == cnt % 2){if (null == oddhead){oddhead = p; oddtail = p;}else{oddtail.next = p;oddtail = oddtail.next;}}else{if (null == evenhead){evenhead = p; eventail = p;}else{eventail.next = p;eventail = eventail.next;}}p = p.next;cnt++;}if (eventail != null) eventail.next = null;oddtail.next = evenhead;return oddhead;}
}

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