原题链接在这里:https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/description/

题目:

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

Example:

Input:
1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL Output:
1-2-3-7-8-11-12-9-10-4-5-6-NULL

Explanation for the above example:

Given the following multilevel doubly linked list:

We should return the following flattened doubly linked list:

题解:

如果当前点cur 没有child, 直接跳到cur.next 进行下次计算.

如果cur 有child, 目标是把cur.child这个level提到cur这个level上. 至于cur.child 这个level上有没有点有child 先不管.

做法就是cur.child 一直只按next找到tail, 然后这一节插在cur 和 cur.next之间, cur再跳到更新的cur.next上.

Time Complexity: O(n). n是所有点的个数, 每个点只走过constant次数.

Space: O(1).

AC Java:

 /*
// Definition for a Node.
class Node {
public int val;
public Node prev;
public Node next;
public Node child; public Node() {} public Node(int _val,Node _prev,Node _next,Node _child) {
val = _val;
prev = _prev;
next = _next;
child = _child;
}
};
*/
class Solution {
public Node flatten(Node head) {
if(head == null){
return head;
} Node cur = head;
while(cur != null){
if(cur.child == null){
cur = cur.next;
continue;
} Node child = cur.child;
Node childTail = child;
while(childTail.next != null){
childTail = childTail.next;
} cur.child = null;
child.prev = cur;
childTail.next = cur.next;
if(cur.next != null){
cur.next.prev = childTail;
}
cur.next = child;
cur = cur.next;
} return head;
}
}

类似Flatten Binary Tree to Linked List.

LeetCode 430. Flatten a Multilevel Doubly Linked List的更多相关文章

  1. 【LeetCode】430. Flatten a Multilevel Doubly Linked List 解题报告(Python)

    [LeetCode]430. Flatten a Multilevel Doubly Linked List 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: ...

  2. LeetCode 430. Faltten a Multilevel Doubly Linked List

    题目链接:LeetCode 430. Faltten a Multilevel Doubly Linked List class Node { public: int val = NULL; Node ...

  3. [LC] 430. Flatten a Multilevel Doubly Linked List

    You are given a doubly linked list which in addition to the next and previous pointers, it could hav ...

  4. 430. Flatten a Multilevel Doubly Linked List

    /* // Definition for a Node. class Node { public: int val = NULL; Node* prev = NULL; Node* next = NU ...

  5. [LeetCode] Flatten a Multilevel Doubly Linked List 压平一个多层的双向链表

    You are given a doubly linked list which in addition to the next and previous pointers, it could hav ...

  6. LeetCode 430:扁平化多级双向链表 Flatten a Multilevel Doubly Linked List

    您将获得一个双向链表,除了下一个和前一个指针之外,它还有一个子指针,可能指向单独的双向链表.这些子列表可能有一个或多个自己的子项,依此类推,生成多级数据结构,如下面的示例所示. 扁平化列表,使所有结点 ...

  7. [LeetCode] 114. Flatten Binary Tree to Linked List 将二叉树展开成链表

    Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 T ...

  8. LeetCode 114| Flatten Binary Tree to Linked List(二叉树转化成链表)

    题目 给定一个二叉树,原地将它展开为链表. 例如,给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解析 通过递归实现:可以用先序遍历, ...

  9. [LeetCode] 114. Flatten Binary Tree to Linked List 将二叉树展平为链表

    Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 ...

随机推荐

  1. SQL使用CASE 语句

    CASE 语句可以在SELECT 子句和ORDER BY 子句中使用 CASE语句分为两种Case Simple Expression and Case Search Expression Case ...

  2. POJ1247-Magnificent Meatballs

    http://poj.org/problem?id=1247 Magnificent Meatballs Time Limit: 1000MS   Memory Limit: 10000K Total ...

  3. ES5下的React

    按照官方推荐的思路,React使用标准的ES6标准的语法.比如说创建一个类: class Greeting extends React.Component { render() { return &l ...

  4. 二十三 Python分布式爬虫打造搜索引擎Scrapy精讲—craw母版l创建自动爬虫文件—以及 scrapy item loader机制

    用命令创建自动爬虫文件 创建爬虫文件是根据scrapy的母版来创建爬虫文件的 scrapy genspider -l  查看scrapy创建爬虫文件可用的母版 Available templates: ...

  5. UVA-11324 The Largest Clique (强连通+DP)

    题目大意:在一张无向图中,最大的节点集使得集合内任意两个节点都能到达对方. 题目分析:找出所有的强连通分量,将每一个分量视作大节点,则原图变成了一张DAG.将每个分量中的节点个数作为节点权值,题目便转 ...

  6. [nodejs]npm国内npm安装nodejs modules终极解决方案

    此方案用于设置代理和修改镜像地址都不能解决问题使用 1.npm root 确认node模块的根文件夹,全局要加-g. osx同样是此命令,先清除缓存. npm cache clean C:\Users ...

  7. RDP协议暴力破解

    真实案例|RDP协议暴力破解卷土重来! 作者:aqniu星期六, 七月 2, 20160   背景 RDP(Remote Desktop Protocol)称为“远程桌面登录协议”,即当某台计算机开启 ...

  8. Spring源码解析-IOC容器的实现

    1.IOC容器是什么? IOC(Inversion of Control)控制反转:本来是由应用程序管理的对象之间的依赖关系,现在交给了容器管理,这就叫控制反转,即交给了IOC容器,Spring的IO ...

  9. uva 12356 Army Buddies 树状数组解法 树状数组求加和恰为k的最小项号 难度:1

    Nlogonia is fighting a ruthless war against the neighboring country of Cubiconia. The Chief General ...

  10. hdu4115

    题解: 2-sat 对于bob出的每一张牌,alice显然只有两种选择 然后对于每一个限制,连边 判断是否可行 代码: #include<cstdio> #include<cmath ...