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[LeetCode] 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 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]

Constraints:

Number of Nodes will not exceed 1000.
1 <= Node.val <= 10^5

扁平化多级双向链表。

题目即是题意，请你将这个多维的链表转化成一个一维的链表。

思路还是像遍历一般链表一样去遍历，当遇到child node的时候，看一下当前节点是否有next node，若有，需要用stack存下来，然后将当前node的next指针指向child，也将child的prev指针指向cur，最后记得将当前节点的child指针置为null。这样起码可以把多维的链表先变成一维的。此时此刻应该是会遍历到链表的尾部了，此时看一下stack是否为空，如果不为空，开始往外弹出节点，接到cur后面，同时记得将cur后边这个节点的prev指针再指向cur，这样才叫双向链表。

时间O(n)

空间O(n) - stack

Java实现

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4     public int val;
 5     public Node prev;
 6     public Node next;
 7     public Node child;
 8 };
 9 */
10 
11 class Solution {
12     public Node flatten(Node head) {
13         // corner case
14         if (head == null) {
15             return head;
16         }
17         Deque<Node> stack = new ArrayDeque<>();
18         Node cur = head;
19         while (cur != null) {
20             if (cur.child != null) {
21                 if (cur.next != null) {
22                     stack.push(cur.next);
23                 }
24                 cur.next = cur.child;
25                 if (cur.next != null) {
26                     cur.next.prev = cur;
27                 }
28                 cur.child = null;
29             } else if (cur.next == null && !stack.isEmpty()) {
30                 cur.next = stack.pop();
31                 if (cur.next != null) {
32                     cur.next.prev = cur;
33                 }
34             }
35             cur = cur.next;
36         }
37         return head;
38     }
39 }