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 ...
给定一个链表,将奇数的节点放在前面,将偶数的节点放在后面。代码如下:
/** * 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 oddNode = head; ListNode evenNode = head.next; ListNode helper = evenNode; while(evenNode != null && evenNode.next != null) { oddNode.next = evenNode.next; evenNode.next = evenNode.next.next; evenNode = evenNode.next; oddNode = oddNode.next; } oddNode.next = helper; return head; } }