LeetCode 430. Flatten a Multilevel Doubly Linked List

原题链接在这里：https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/description/

题目：

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:

题解：

如果当前点cur 没有child, 直接跳到cur.next 进行下次计算.

如果cur 有child, 目标是把cur.child这个level提到cur这个level上. 至于cur.child 这个level上有没有点有child 先不管. 

做法就是cur.child 一直只按next找到tail, 然后这一节插在cur 和 cur.next之间, cur再跳到更新的cur.next上.

Time Complexity: O(n). n是所有点的个数, 每个点只走过constant次数.

Space: O(1).

AC Java:

/*
// Definition for a Node.
class Node {
    public int val;
    public Node prev;
    public Node next;
    public Node child;

    public Node() {}

    public Node(int _val,Node _prev,Node _next,Node _child) {
        val = _val;
        prev = _prev;
        next = _next;
        child = _child;
    }
};
*/
class Solution {
    public Node flatten(Node head) {
        if(head == null){
            return head;
        }
        
        Node cur = head;
        while(cur != null){
            if(cur.child == null){
                cur = cur.next;
                continue;
            }
            
            Node child = cur.child;
            Node childTail = child;
            while(childTail.next != null){
                childTail = childTail.next;
            }
            
            cur.child = null;
            child.prev = cur;
            childTail.next = cur.next;
            if(cur.next != null){
                cur.next.prev = childTail;
            }
            cur.next = child;
            cur = cur.next;
        }
        
        return head;
    }
}

类似Flatten Binary Tree to Linked List.