0430. Flatten a Multilevel Doubly Linked List (M)

Posted 2020-12-17 mapoos

Flatten a Multilevel Doubly Linked List (M)

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 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

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题意

给双向链表多加一个指针域child，指向下一层的链表。要求将给定的多层链表压平成一个链表。

思路

DFS处理，从下往上合并每一层；也可以直接用迭代从上往下合并每一层。

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代码实现

Java

递归

class Solution {
    public Node flatten(Node head) {
        dfs(head);
        return head;
    }

  	// dfs返回压平后的当前链表的尾结点
    private Node dfs(Node head) {
        Node p = head;
        Node last = null;
        while (p != null) {
            if (p.next == null) {
                last = p;
            }
            if (p.child != null) {
                Node childLast = dfs(p.child);
                Node next = p.next;
                p.next = p.child;
                p.child.prev = p;
                p.child = null;
                childLast.next = next;
                if (next != null) {
                    next.prev = childLast;
                }
                p = childLast;
            } else {
                p = p.next;
            }
        }
        return last;
    }
}

迭代

class Solution {
    public Node flatten(Node head) {
        Node p = head;
        while (p != null) {
            Node next = p.next;
            if (p.child != null) {
                Node q = p.child;
                while (q.next != null) {
                    q = q.next;
                }
                p.next = p.child;
                p.child.prev = p;
                p.child = null;
                q.next = next;
                if (next != null) {
                    next.prev = q;
                }
            }
            p = p.next;
        }
        return head;
    }
}