Description
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]
思路
- 使用循环，从头开始遍历，将子链看成一个整体，将整体向上移动一层。
- 继续向后遍历，如果遇到子链，将其看作一层，将其整体向上移动，如此循环下去，直至结束。
# -*- coding: utf-8 -*- # @Author: 何睿 # @Create Date: 2019-12-29 10:57:03 # @Last Modified by: 何睿 # @Last Modified time: 2019-12-29 11:05:17 class Node: def __init__(self, val, prev, next, child): self.val = val self.prev = prev self.next = next self.child = child class Solution: def flatten(self, head: 'Node') -> 'Node': current = head while current: if current.child: _next = current.next # 当前节点的后一个节点 last = current.child # 当前节点的自节点 while last.next: last = last.next # 如果有子节点，找到子节点的最后一个节点 current.next = current.child current.next.prev = current # 将自节点向上提高一层 current.child = None last.next = _next if _next: _next.prev = last current = current.next return head
源代码文件在 这里 。
