标签:appear addition code nat sig previous ext one alt
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 1:
Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12] Output: [1,2,3,7,8,11,12,9,10,4,5,6] Explanation: The multilevel linked list in the input is as follows:
After flattening the multilevel linked list it becomes:
Example 2:
Input: head = [1,2,null,3] Output: [1,3,2] Explanation: The input multilevel linked list is as follows: 1---2---NULL | 3---NULL
Example 3:
Input: head = [] Output: []
How multilevel linked list is represented in test case:
We use the multilevel linked list from Example 1 above:
1---2---3---4---5---6--NULL | 7---8---9---10--NULL | 11--12--NULL
The serialization of each level is as follows:
[1,2,3,4,5,6,null] [7,8,9,10,null] [11,12,null]
To serialize all levels together we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:
[1,2,3,4,5,6,null] [null,null,7,8,9,10,null] [null,11,12,null]
Merging the serialization of each level and removing trailing nulls we obtain:
[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
class Solution { public Node flatten(Node head) { if( head == null) return head; // Pointer Node p = head; while( p!= null) { /* CASE 1: if no child, proceed */ if( p.child == null ) { p = p.next; continue; } /* CASE 2: got child, find the tail of the child and link it to p.next */ Node temp = p.child; // Find the tail of the child while( temp.next != null ) temp = temp.next; // Connect tail with p.next, if it is not null temp.next = p.next; if( p.next != null ) p.next.prev = temp; // Connect p with p.child, and remove p.child p.next = p.child; p.child.prev = p; p.child = null; } return head; } }
430. Flatten a Multilevel Doubly Linked List
标签:appear addition code nat sig previous ext one alt
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/13282219.html