Add solution #1019

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,093 LeetCode solutions in JavaScript
+# 1,094 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -827,6 +827,7 @@
 1016|[Binary String With Substrings Representing 1 To N](./solutions/1016-binary-string-with-substrings-representing-1-to-n.js)|Medium|
 1017|[Convert to Base -2](./solutions/1017-convert-to-base-2.js)|Medium|
 1018|[Binary Prefix Divisible By 5](./solutions/1018-binary-prefix-divisible-by-5.js)|Easy|
+1019|[Next Greater Node In Linked List](./solutions/1019-next-greater-node-in-linked-list.js)|Medium|
 1022|[Sum of Root To Leaf Binary Numbers](./solutions/1022-sum-of-root-to-leaf-binary-numbers.js)|Easy|
 1023|[Camelcase Matching](./solutions/1023-camelcase-matching.js)|Medium|
 1028|[Recover a Tree From Preorder Traversal](./solutions/1028-recover-a-tree-from-preorder-traversal.js)|Hard|

solutions/1019-next-greater-node-in-linked-list.js

@@ -0,0 +1,46 @@
+/**
+ * 1019. Next Greater Node In Linked List
+ * https://leetcode.com/problems/next-greater-node-in-linked-list/
+ * Difficulty: Medium
+ *
+ * You are given the head of a linked list with n nodes.
+ *
+ * For each node in the list, find the value of the next greater node. That is, for each node,
+ * find the value of the first node that is next to it and has a strictly larger value than it.
+ *
+ * Return an integer array answer where answer[i] is the value of the next greater node of the
+ * ith node (1-indexed). If the ith node does not have a next greater node, set answer[i] = 0.
+ */
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {number[]}
+ */
+var nextLargerNodes = function(head) {
+  const values = [];
+  let current = head;
+
+  while (current) {
+    values.push(current.val);
+    current = current.next;
+  }
+
+  const result = new Array(values.length).fill(0);
+  const stack = [];
+
+  for (let i = 0; i < values.length; i++) {
+    while (stack.length && values[stack[stack.length - 1]] < values[i]) {
+      result[stack.pop()] = values[i];
+    }
+    stack.push(i);
+  }
+
+  return result;
+};