- 重排链表
难度中等
给定一个单链表 L 的头节点 head ,单链表 L 表示为:
L0 → L1 → … → Ln - 1 → Ln
请将其重新排列后变为:
L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …
不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例 1:
输入:head = [1,2,3,4]
输出:[1,4,2,3]
示例 2:
输入:head = [1,2,3,4,5]
输出:[1,5,2,4,3]
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public void reorderList(ListNode head) {
if(head == null || head.next == null || head.next.next == null){
return;
}
ListNode pre = null ;
ListNode cur = head;
while(cur.next != null){
pre = cur;
cur = cur.next;
}
ListNode sec = head.next;
head.next = cur;
cur.next = sec;
pre.next = null;
reorderList(sec);
}
}