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;
}
}