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leetcode_java/src/main/java/LeetCodeJava/LinkedList/FlattenMultilevelDoublyLinkedList.java
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| package LeetCodeJava.LinkedList; | ||
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| // https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/description/ | ||
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| import java.util.LinkedList; | ||
| import java.util.Queue; | ||
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| /** | ||
| * 430. Flatten a Multilevel Doubly Linked List | ||
| * Solved | ||
| * Medium | ||
| * Topics | ||
| * Companies | ||
| * 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. | ||
| * <p> | ||
| * 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 after curr and before curr.next in the flattened list. | ||
| * <p> | ||
| * Return the head of the flattened list. The nodes in the list must have all of their child pointers set to null. | ||
| * <p> | ||
| * <p> | ||
| * <p> | ||
| * Example 1: | ||
| * <p> | ||
| * <p> | ||
| * 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: | ||
| * <p> | ||
| * Example 2: | ||
| * <p> | ||
| * <p> | ||
| * 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: | ||
| * <p> | ||
| * Example 3: | ||
| * <p> | ||
| * Input: head = [] | ||
| * Output: [] | ||
| * Explanation: There could be empty list in the input. | ||
| * <p> | ||
| * <p> | ||
| * Constraints: | ||
| * <p> | ||
| * The number of Nodes will not exceed 1000. | ||
| * 1 <= Node.val <= 105 | ||
| * <p> | ||
| * <p> | ||
| * How the multilevel linked list is represented in test cases: | ||
| * <p> | ||
| * We use the multilevel linked list from Example 1 above: | ||
| * <p> | ||
| * 1---2---3---4---5---6--NULL | ||
| * | | ||
| * 7---8---9---10--NULL | ||
| * | | ||
| * 11--12--NULL | ||
| * The serialization of each level is as follows: | ||
| * <p> | ||
| * [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: | ||
| * <p> | ||
| * [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: | ||
| * <p> | ||
| * [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12] | ||
| */ | ||
| public class FlattenMultilevelDoublyLinkedList { | ||
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| /* | ||
| // Definition for a Node. | ||
| class Node { | ||
| public int val; | ||
| public Node prev; | ||
| public Node next; | ||
| public Node child; | ||
| }; | ||
| */ | ||
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| // V0 | ||
| // TODO : implement | ||
| // public Node flatten(Node head) { | ||
| // | ||
| // } | ||
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| // V1 | ||
| // https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/solutions/5667355/easy-java-solution/ | ||
| class Node { | ||
| public int val; | ||
| public Node prev; | ||
| public Node next; | ||
| public Node child; | ||
| }; | ||
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| private Queue<Node> store = new LinkedList<>(); | ||
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| public void helper(Node head) { | ||
| Node temp; | ||
| while (head != null) { | ||
| temp = head.next; | ||
| head.next = null; | ||
| head.prev = null; | ||
| store.offer(head); | ||
| if (head.child != null) | ||
| helper(head.child); | ||
| head.child = null; | ||
| head = temp; | ||
| } | ||
| } | ||
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| public Node flatten_1(Node head) { | ||
| helper(head); | ||
| if (store.peek() == null) | ||
| return head; | ||
| Node retval = store.poll(); | ||
| Node first = retval; | ||
| while (store.peek() != null) { | ||
| Node second = store.poll(); | ||
| first.next = second; | ||
| second.prev = first; | ||
| first = second; | ||
| } | ||
| first.next = null; | ||
| return retval; | ||
| } | ||
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| // V2 | ||
| // IDEA : LINKED LIST + CUSTOM CLASS | ||
| // https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/solutions/4031041/single-pass-solution-using-custom-class/ | ||
| class lc430Helper { | ||
| Node head; | ||
| Node tail; | ||
| } | ||
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| public Node flatten_2(Node head) { | ||
| return util1(head).head; | ||
| } | ||
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| public lc430Helper util1(Node head) { | ||
| Node dummy = new Node(); | ||
| dummy.next = head; | ||
| Node temp = dummy; | ||
| while (temp.next != null) { | ||
| temp = temp.next; | ||
| if (temp.child == null) { | ||
| continue; | ||
| } else { | ||
| lc430Helper bottom = util1(temp.child); | ||
| Node temp2 = temp.next; | ||
| temp.child = null; | ||
| temp.next = bottom.head; | ||
| bottom.head.prev = temp; | ||
| bottom.tail.next = temp2; | ||
| if (temp2 != null) { | ||
| temp2.prev = bottom.tail; | ||
| } | ||
| temp = bottom.tail; | ||
| } | ||
| } | ||
| lc430Helper ans = new lc430Helper(); | ||
| dummy.next = null; | ||
| ans.head = head; | ||
| ans.tail = temp; | ||
| return ans; | ||
| } | ||
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| // V3 | ||
| // https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/solutions/5328452/easy-to-understand-best-solution/ | ||
| public Node flatten_3(Node head) { | ||
| Node temp = head; | ||
| while(temp != null){ | ||
| Node list1Tail = temp; | ||
| Node list3Head = temp.next; | ||
| // if the node has child node then append all its node in between | ||
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| if(temp.child != null){ | ||
| // we are assuming that recursion will give us flatten output, so we just need to adjust the pointers | ||
| Node list2Head = flatten_3(temp.child); | ||
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| // find list2 tail | ||
| Node list2Tail = list2Head; | ||
| while(list2Tail.next != null){ | ||
| list2Tail = list2Tail.next; | ||
| } | ||
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| // attach the lists | ||
| list1Tail.next = list2Head; | ||
| list2Head.prev = list1Tail; | ||
| list2Tail.next = list3Head; | ||
| if(list3Head != null) | ||
| list3Head.prev = list2Tail; | ||
| temp.child = null; | ||
| } | ||
| temp = temp.next; | ||
| } | ||
| return head; | ||
| } | ||
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| } |