430. 扁平化多级双向链表
430. 扁平化多级双向链表
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链表
双向链表
🔗 力扣
LeetCode
题目
You are given a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. 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.
Given the head
of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let curr
be a node with a child list. The nodes in the child list should appear aftercurr
and before curr.next
in the flattened list.
Return thehead
_of the flattened list. The nodes in the list must have all of their child pointers set to _null
.
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 shown.
After flattening the multilevel linked list it becomes:
Example 2:
Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:
Example 3:
Input: head = []
Output: []
Explanation: There could be empty list in the input.
Constraints:
- The number of Nodes will not exceed
1000
. 1 <= Node.val <= 10^5
How the multilevel linked list is represented in test cases:
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]
题目大意
你会得到一个双链表,其中包含的节点有一个下一个指针、一个前一个指针和一个额外的 子指针 。这个子指针可能指向一个单独的双向链表,也包含这些特殊的节点。这些子列表可以有一个或多个自己的子列表,以此类推,以生成如下面的示例所示的 多层数据结构 。
给定链表的头节点 head
,将链表 扁平化 ,以便所有节点都出现在单层双链表中。让 curr
是一个带有子列表的节点。子列表中的节点应该出现在扁平化列表中的 curr
之后 和 curr.next
之前 。
返回 扁平列表的 head
。列表中的节点必须将其 所有 子指针设置为 null
。
解题思路
- 使用递归遍历多级链表,如果某个节点有子链表,则对子链表进行递归处理。
- 在递归处理过程中,将当前节点的
next
指针指向递归处理后的子链表,同时将子链表的prev
指针指向当前节点。 - 最后,将当前节点的
child
指针置为null
。
复杂度分析
- 时间复杂度:
O(n)
,其中n
是链表的节点总数。这个算法递归地处理了多级链表中的每个节点,每个节点都只处理一次。 - 空间复杂度:
O(1)
。
代码
/**
* @param {Node} head
* @return {Node}
*/
var flatten = function (head) {
if (!head) return null;
const next = head.next;
const child = head.child;
// 处理当前节点的子链表
const flattenChild = flatten(child);
// 连接当前节点和递归处理后的子链表
if (flattenChild) {
head.next = flattenChild;
flattenChild.prev = head;
head.child = null;
}
// 递归处理后续节点
const flattenNext = flatten(next);
// 连接递归处理后的子链表和后续节点
if (flattenNext) {
const last = findLast(flattenChild) || head;
last.next = flattenNext;
flattenNext.prev = last;
}
return head;
};
// 辅助函数:找到链表的最后一个节点
var findLast = function (head) {
while (head && head.next) {
head = head.next;
}
return head;
};
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