430.Flatten a Multilevel Doubly Linked List¶
Tags: Medium
Linked List
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Links: https://leetcode.com/problems/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
很明显对于某个节点存在child
的时候,利用递归去展开子节点,假设某个child
展开的结果是front
,我们还需要一个指针指向尾节点,先当与把一段用front
和 end
给包含起来。注意一个点,当访问head -> next
的时候,一定要注意检验其是否存在,如果不存在就不需要建立prev
指针了。
所有节点最多遍历两次,时间复杂度O(n)。
/*
// Definition for a Node.
class Node {
public:
int val;
Node* prev;
Node* next;
Node* child;
};
*/
class Solution {
public:
Node* flatten(Node* head) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
if (!head) return NULL;
Node * dummy = new Node(-1);
dummy -> next = head;
while (head) {
if (head -> child) {
Node * front = flatten(head -> child);
Node * end = getEnd(front);
head -> child = NULL;
if (head -> next) head -> next -> prev = end;
end -> next = head -> next;
head -> next = front;
front -> prev = head;
head = end -> next;
}
else {
head = head -> next;
}
}
Node *res = dummy -> next;
delete dummy; dummy = NULL;
return res;
}
Node *getEnd(Node *head)
{
if (!head) return NULL;
while (head -> next) {
head = head -> next;
}
return head;
}
};