题意:
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。
题解:
可以发现重新排列的链表,是原链表前半段和反转后的后半段交替
- 使用快慢指针找到链表的中间节点,将链表从中间节点断开,保证链表前半段长度不小于后半段
- 将后半段的链表反转
- 重新合并两个链表
点击查看代码
/**
* 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) {
ListNode dummy = new ListNode(0, head);
ListNode slow = dummy, fast = dummy;
while(fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode fir = head, sec = slow.next;
sec = reverse(sec);
while(sec != null) {
ListNode nextSec = sec.next;
sec.next = fir.next;
fir.next = sec;
fir = sec.next;
sec = nextSec;
}
fir.next = null;
}
public ListNode reverse(ListNode head) {
if(head == null || head.next == null) return head;
ListNode res = reverse(head.next);
head.next.next = head;
head.next = null;
return res;
}
}