【LeetCode】 143. Reorder List 重排链表（Medium）（JAVA）每日一题

【LeetCode】 143. Reorder List 重排链表（Medium）（JAVA）

题目地址: https://leetcode.com/problems/reorder-list/

题目描述：

Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

题目大意

给定一个单链表 L：L0→L1→…→Ln-1→Ln ， 将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你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。

解题方法

1、快慢指针找出中间节点 2、中间节点之后的所有元素翻转 3、从头开始遍历一个个元素接起来

class Solution { public void reorderList(ListNode head) { if (head == null || head.next == null) return; ListNode fast = head.next; ListNode slow = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } ListNode start = slow.next; slow.next = null; ListNode reverse = null; while (start != null) { ListNode next = start.next; start.next = reverse; reverse = start; start = next; } start = head; while (start != null && reverse != null) { ListNode next = start.next; start.next = reverse; reverse = reverse.next; start.next.next = next; start = next; } } }

执行耗时:1 ms,击败了100.00% 的Java用户 内存消耗:40.5 MB,击败了99.74% 的Java用户