328. 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.
Example 1:
Input: 1->2->3->4->5->NULL Output: 1->3->5->2->4->NULL
Example 2:
Input: 2->1->3->5->6->4->7->NULL Output: 2->3->6->7->1->5->4->NULL
Note:
思路:
题目意思将链表中的奇数位节点放在链表前面,偶数位节点放在链表后面,简单的实现题,注意的点主要是不要产生断链(记录号偶数节点的起始位置)和循环结束条件,避免出现空针等情况。
代码:
java:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode oddEvenList(ListNode head) {
if (head == null || head.next == null) return head;
ListNode odd = head, even = head.next, evenStart = even;
while (even != null && even.next != null) {
odd.next = odd.next.next;
even.next = even.next.next;
odd = odd.next;
even = even.next;
}
odd.next = evenStart;
return head;
}
}