leetcode328~Odd Even Linked List

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.

大致翻译：给定一个链表，将偶数位置的节点移动至奇数位置节点的后面，要求时间复杂度为O(1)。

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 …

思路：对于链表类题目，要求时间复杂度为O(1)的常见做法就是双指针~

public class OddEvenLinkedList {
	public ListNode oddEvenList(ListNode head) {
		if(head==null || head.next==null ||head.next.next==null) return head;
		//奇偶链表
		ListNode oddhead = head;
		ListNode evenhead = head.next;
		ListNode oddcur = oddhead;
		ListNode evencur = evenhead;
		while(evencur!=null && evencur.next!=null) {
			oddcur.next = evencur.next;
			evencur.next = evencur.next.next;
			oddcur = oddcur.next;
			evencur = evencur.next;
		}
		oddcur.next = evenhead;
		return oddhead;
	}
}