LeetCode 430. Flatten a Multilevel Doubly Linked List

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.