430. Flatten a Multilevel Doubly Linked List.md

430. Flatten a Multilevel Doubly Linked List

Difficulty : Medium

Related Topics : LinkedList、DFS

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leetcode Daily Challenge on July 10th, 2020.

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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 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 as follows:

After flattening the multilevel linked list it becomes:

Example 2:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation:

The input multilevel linked list is as follows:

  1---2---NULL
  |
  3---NULL

Example 3:

Input: head = []
Output: []

How multilevel linked list is represented in test case:

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]

Constraints:

• Number of Nodes will not exceed 1000.
• 1 <= Node.val <= 10^5

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Solution

• Java
  • mine

    • Recursive Runtime: 1 ms, faster than 14.60%, Memory Usage: 37.3 MB, less than 85.00% of Java online submissions

      //O(N)time
      //O(1)space
      public Node flatten(Node head) {
          if (head == null) {
              return head;
          }
          Node res = new Node();
          Node t = res;
          Node next;
          while (head != null) {
              next = head.next;
              Node child = head.child;
              head.child = null;
              head.next = null;
              t.next = head;
              t.next.prev = t;
              t = t.next;
              if (child != null) {
                  Node cRes = flatten(child);
                  t.next = cRes;
                  cRes.prev = t;
                  while (t.next != null) {
                      t = t.next;
                  }
              }
              head = next;
          }
          res.next.prev = null;
          return res.next;
      }
      

    • DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 37.3 MB, less than 89.09% of Java online submissions

      // O(N)time 
      // O(C)space  C is the count of node has child
      public Node flatten(Node head) {
          LinkedList<Node> list = new LinkedList<>();
          Node res = head;
          Node t = res;
          while (t != null) {
              if (t.child != null) {
                  if (t.next != null) {
                      list.add(t.next);
                  }
                  t.child.prev = t;
                  t.next = t.child;
                  t.child = null;
              } else {
                  if (t.next == null) {
                      if (list.isEmpty()) {
                          break;
                      }
                      t.next = list.removeLast();
                  }
                  t.next.prev = t;
              }
              t = t.next;
          }
          return res;
      }
      

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• the most votes

  • Runtime: 0 ms, faster than 100.00%, Memory Usage: 37.7 MB, less than 75.00% of Java online submissions

    // O(N)time 
    // O(1)space
    public Node flatten(Node head) {
        if (head == null) return head;
        // Pointer
        Node p = head;
        while (p != null) {
            /* CASE 1: if no child, proceed */
            if (p.child == null) {
                p = p.next;
                continue;
            }
            /* CASE 2: got child, find the tail of the child and link it to p.next */
            Node temp = p.child;
            // Find the tail of the child
            while (temp.next != null)
                temp = temp.next;
            // Connect tail with p.next, if it is not null
            temp.next = p.next;
            if (p.next != null) p.next.prev = temp;
            // Connect p with p.child, and remove p.child
            p.next = p.child;
            p.child.prev = p;
            p.child = null;
        }
        return head;
    }
    

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