LeetCode: 430. Flatten a Multilevel Doubly Linked List¶
1、题目描述¶
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:
Input:
1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL
Output:
1-2-3-7-8-11-12-9-10-4-5-6-NULL
Explanation for the above example:
Given the following multilevel doubly linked list:
We should return the following flattened doubly linked list:
2、解题思路¶
可以使用深度优先搜索，也可以不是用，首先来剖析一下问题
这道题与前面114题很相似，如果当前的节点有子节点，该如何做呢？
将剩下的节点，挂到子节点的末尾，然后，将当前节点的next指向子节点，子节点的prev指向当前节点即可，然后节点后移，继续判断
class Solution(object): def flatten(self, head): """ :type head: Node :rtype: Node """ ans = head while head: if head.child: head_next = head.next temp = head.child while temp.next: temp = temp.next temp.next = head_next head_next.prev = temp head.next = head.child head.child.prev = head head.child = None head = head.next return ans
思路很简单，但是通过了19个测试用了，下一个超时了，因此，要考虑如何提高效率
考虑在展开的过程中，不直接将下一个元素挂到子列的尾部，而是保存到栈中，等到遍历到最后的时候，就将栈中的元素取出来，接到尾部，然后继续判断即可
""" # Definition for a Node. class Node(object): def __init__(self, val, prev, next, child): self.val = val self.prev = prev self.next = next self.child = child """ class Solution(object): def flatten(self, head): """ :type head: Node :rtype: Node """ ans = head if not head: return ans stack = [] prev = None while head or stack: if head: prev = head if head.child: if head.next: stack.append(head.next) head.next = head.child head.child.prev = head head.child = None head = head.next else: temp = stack.pop() prev.next = temp temp.prev = prev head = temp return ans
这样，仅需要从头到尾遍历一遍即可得出结果