Add solution #430

Author: JoshCrozier

README.md

@@ -341,6 +341,7 @@
 424|[Longest Repeating Character Replacement](./0424-longest-repeating-character-replacement.js)|Medium|
 427|[Construct Quad Tree](./0427-construct-quad-tree.js)|Medium|
 429|[N-ary Tree Level Order Traversal](./0429-n-ary-tree-level-order-traversal.js)|Medium|
+430|[Flatten a Multilevel Doubly Linked List](./0430-flatten-a-multilevel-doubly-linked-list.js)|Medium|
 434|[Number of Segments in a String](./0434-number-of-segments-in-a-string.js)|Easy|
 435|[Non-overlapping Intervals](./0435-non-overlapping-intervals.js)|Medium|
 437|[Path Sum III](./0437-path-sum-iii.js)|Medium|

solutions/0430-flatten-a-multilevel-doubly-linked-list.js

@@ -0,0 +1,61 @@
+/**
+ * 430. Flatten a Multilevel Doubly Linked List
+ * https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/
+ * Difficulty: Medium
+ *
+ * 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.
+ *
+ * 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.
+ *
+ * Return the head of the flattened list. The nodes in the list must have all of their child
+ * pointers set to null.
+ */
+
+/**
+ * // Definition for a _Node.
+ * function _Node(val,prev,next,child) {
+ *    this.val = val;
+ *    this.prev = prev;
+ *    this.next = next;
+ *    this.child = child;
+ * };
+ */
+
+/**
+ * @param {_Node} head
+ * @return {_Node}
+ */
+var flatten = function(head) {
+  if (!head) return null;
+
+  let current = head;
+  while (current) {
+    if (!current.child) {
+      current = current.next;
+    } else {
+      const { child, next } = current;
+      current.child = null;
+      current.next = child;
+      child.prev = current;
+
+      let tail = child;
+      while (tail.next) {
+        tail = tail.next;
+      }
+
+      tail.next = next;
+      if (next) {
+        next.prev = tail;
+      }
+      current = current.next;
+    }
+  }
+
+  return head;
+};