Reorder List

最新推荐文章于 2022-10-08 18:49:12 发布

cgdong2014

最新推荐文章于 2022-10-08 18:49:12 发布

阅读量174

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode LinkedList

本文链接：https://blog.youkuaiyun.com/cgdong2014/article/details/39430399

LeetCode 同时被 2 个专栏收录

139 篇文章

订阅专栏

LinkedList

11 篇文章

订阅专栏

本文详细阐述了如何通过双指针、反转链表和交错合并的方法，将单链表重新排列为奇数节点和偶数节点交替出现的形式。重点讨论了算法实现的关键步骤和注意事项，确保操作不改变节点值且在原地完成。

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

You must do this in-place without altering the nodes' values.

For example,

Given {1,2,3,4}, reorder it to {1,4,2,3}.

思路: 1. Set a slow pointer and fast pointer to find the middle of the link

2. Reverse the rear half list.

3. Merge the two list. One by one.

易错点：注意分开后的两个list 一定要断开连接， slow.next = null !!!! 还有就是判断循环结束用 fast.next != null && fast.next.next != null

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if(head == null)
            return;
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        fast = slow.next;// The rear half linked list
        slow.next = null;
        fast = reverseLinkList(fast);//Reverse the list
        slow = head;
        while(fast != null){
            ListNode cur = fast;
            fast = fast.next;
            cur.next = slow.next;
            slow.next = cur;
            slow = cur.next;
        }
    }
    
    private ListNode reverseLinkList(ListNode head){
        if(head == null)
            return head;
        ListNode newHead = new ListNode(0);
        while(head != null){
            ListNode cur = head;
            head = head.next;
            cur.next = newHead.next;
            //newHead.next = cur.next;
            newHead.next = cur;
        }
        return newHead.next;
    }
}