Add solution #148

Author: JoshCrozier

README.md

@@ -117,6 +117,7 @@
 143|[Reorder List](./0143-reorder-list.js)|Medium|
 144|[Binary Tree Preorder Traversal](./0144-binary-tree-preorder-traversal.js)|Easy|
 145|[Binary Tree Postorder Traversal](./0145-binary-tree-postorder-traversal.js)|Easy|
+148|[Sort List](./0148-sort-list.js)|Medium|
 149|[Max Points on a Line](./0149-max-points-on-a-line.js)|Hard|
 151|[Reverse Words in a String](./0151-reverse-words-in-a-string.js)|Medium|
 152|[Maximum Product Subarray](./0152-maximum-product-subarray.js)|Medium|

solutions/0148-sort-list.js

@@ -0,0 +1,49 @@
+/**
+ * 148. Sort List
+ * https://leetcode.com/problems/sort-list/
+ * Difficulty: Medium
+ *
+ * Given the head of a linked list, return the list after sorting it in ascending order.
+ */
+
+/**
+ * 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 {ListNode}
+ */
+var sortList = function(head) {
+  if (!head || !head.next) return head;
+  let fast = head;
+  let slow = head;
+  while (fast.next && fast.next.next) {
+    fast = fast.next.next;
+    slow = slow.next;
+  }
+  const middle = slow.next;
+  slow.next = null;
+  return mergeList(sortList(head), sortList(middle));
+};
+
+function mergeList(a, b) {
+  const ordered = new ListNode(-1);
+  let list = ordered;
+
+  while (a && b) {
+    list.next = a.val < b.val ? a : b;
+    list = list.next;
+    if (a.val < b.val) {
+      a = a.next;
+    } else {
+      b = b.next;
+    }
+  }
+
+  list.next = a ?? b;
+  return ordered.next;
+}