LeetCode之旅（15）-Odd Even Linked List

题目描述：

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.

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

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Example:

Given 1->2->3->4->5->NULL,

return 1->3->5->2->4->NULL.

注意：

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 …

思路分析：

给定一个单向的列表，然后将里面的奇数和偶数分开，偶数链接在奇数后面，而且要求是不能增加存储空间，以及在现行时间内解决

可以设置一个a节点，以及一个b节点，a节点指向奇数链条的最后一个，b节点指向偶数链条的最后一个（下一个是奇数了） 。比如1->3->5->2->4->6->7->8->9,那么a = 5，b = 6.

首先是把 a.next->7,b.next->7.next(8),7.next->2,就便成了下面的样子

1->3->5->7->2->4->6->8->9##

然后a和b往后移动，a = 7，b = 8

代码实现：

/**

• 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) return head;
  ListNode a = head, b =head;
  while (b != null) {
  b = b.next;
  if (b==null || b.next==null) break;
  ListNode a_next = a.next, b_next = b.next;
  b.next = b_next.next;
  a.next = b_next;
  b_next.next = a_next;
  a = a.next;
  }
  return head;
  }
  } ##