Reorder List ---LeetCode

来源：互联网发布：淘宝店铺第三方编辑：程序博客网时间：2024/06/11 18:17

https://leetcode.com/problems/reorder-list/

解题思路：
首先题目要求了要 in-place。表明我们不能创建一个新的链表，而只能通过改变指针的指向来重新排序。

一共分为三个步骤解决：

使用 slow 和 fast 指针找到链表中点，将链表分为两半。
将后半段链表反转。
最后将两段链表归并即可。

示意图：

这里写图片描述

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public void reorderList(ListNode head) {        if (head == null || head.next == null) return;        ListNode slow = head;        ListNode fast = head;        while (fast.next != null && fast.next.next != null) {            slow = slow.next;            fast = fast.next.next;        }        ListNode head1 = head;        ListNode head2 = slow.next;        slow.next = null;        head2 = reverse(head2);        head = merge(head1, head2);    }    public ListNode reverse (ListNode head) {        ListNode prev = head;        ListNode curr = head.next;        head.next = null;        while (curr != null) {            ListNode next = curr.next;            curr.next = prev;            prev = curr;            curr = next;        }        return prev;    }    public ListNode merge (ListNode l1, ListNode l2) {              while (l2 != null) {            ListNode t1 = l1.next;            ListNode t2 = l2.next;            l1.next = l2;            l2.next = t1;            l1 = t1;            l2 = t2;        }        return l1;    }}

0 0