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]
Constraints:
- Number of Nodes will not exceed 1000.
1 <= Node.val <= 10^5
Solutions
1. Coding Skill
主要考验的是编程能力，看能不能够 bug free 了！
# Time: O(n)
# Space: O(1)
"""
# Definition for a Node.
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':
if not head:
return head
cur = head
while cur:
# CASE 1: if no child, proceed
if not cur.child:
cur = cur.next
continue
# CASE 2: got child, find the tail of the child and link it to cur.next
tmp = cur.child
# find the tail of the child
while tmp.next:
tmp = tmp.next
# connect tail with cur.next, if it is not None
tmp.next = cur.next
if cur.next:
cur.next.prev = tmp
cur.next = cur.child
cur.child.prev = cur
cur.child = None
return head
# 22/22 cases passed (36 ms)
# Your runtime beats 72.4 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)
2. Stack + two pointers
总有一个先处理 child 后处理 next 的操作，就可以用 stack 来存。
class Solution:
def flatten(self, head: 'Node') -> 'Node':
if not head:
return head
stack = [head]
prev = Node(0)
while stack:
node = stack.pop()
node.prev = prev
prev.next = node
prev = node
if node.next:
stack.append(node.next)
if node.child:
stack.append(node.child)
node.child = None
head.prev = None
return head
# 22/22 cases passed (32 ms)
# Your runtime beats 87.74 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)