题解 | #链表的奇偶重排#
链表的奇偶重排
https://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3
import java.util.*; /* * public class ListNode { * int val; * ListNode next = null; * public ListNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head ListNode类 * @return ListNode类 */ public ListNode oddEvenList (ListNode head) { // write code here if(head == null){ return head; } List<ListNode> jsList = new ArrayList<>(); List<ListNode> osList = new ArrayList<>(); ListNode jsNode = head; ListNode osNode = head.next; //奇数节点入List while (jsNode != null) { jsList.add(jsNode); jsNode = jsNode.next; if(jsNode != null){ jsNode = jsNode.next; } } //偶数节点入List while (osNode != null) { osList.add(osNode); osNode = osNode.next; if(osNode != null){ osNode = osNode.next; } } //定义虚拟节点 ListNode newHeadNode = new ListNode(-1); ListNode tempHead = newHeadNode; for (int i = 0; i < jsList.size(); i++) { newHeadNode.next = jsList.get(i); newHeadNode = newHeadNode.next; } for (int i = 0; i < osList.size(); i++) { newHeadNode.next = osList.get(i); newHeadNode = newHeadNode.next; } //newHeadNode.next = null 防止产生环 newHeadNode.next = null; return tempHead.next; } }