Add solution #86

Author: JoshCrozier

README.md

@@ -74,6 +74,7 @@
 78|[Subsets](./0078-subsets.js)|Medium|
 80|[Remove Duplicates from Sorted Array II](./0080-remove-duplicates-from-sorted-array-ii.js)|Medium|
 83|[Remove Duplicates from Sorted List](./0083-remove-duplicates-from-sorted-list.js)|Easy|
+86|[Partition List](./0086-partition-list.js)|Medium|
 88|[Merge Sorted Array](./0088-merge-sorted-array.js)|Easy|
 90|[Subsets II](./0090-subsets-ii.js)|Medium|
 93|[Restore IP Addresses](./0093-restore-ip-addresses.js)|Medium|

solutions/0086-partition-list.js

@@ -0,0 +1,31 @@
+/**
+ * 86. Partition List
+ * https://leetcode.com/problems/partition-list/
+ * Difficulty: Medium
+ *
+ * Given the head of a linked list and a value x, partition it such that
+ * all nodes less than x come before nodes greater than or equal to x.
+ *
+ * You should preserve the original relative order of the nodes in each
+ * of the two partitions.
+ */
+
+/**
+ * @param {ListNode} head
+ * @param {number} x
+ * @return {ListNode}
+ */
+var partition = function(head, x) {
+  const result = [];
+  const stack = [];
+
+  while (head) {
+    const target = head.val >= x ? result : stack;
+    target.push(head.val);
+    head = head.next;
+  }
+
+  return [...stack, ...result].reverse().reduce((a, b) => {
+    return new ListNode(b, a);
+  }, null);
+};