Add solution #147

Author: JoshCrozier

README.md

@@ -149,6 +149,7 @@
 144|[Binary Tree Preorder Traversal](./0144-binary-tree-preorder-traversal.js)|Easy|
 145|[Binary Tree Postorder Traversal](./0145-binary-tree-postorder-traversal.js)|Easy|
 146|[LRU Cache](./0146-lru-cache.js)|Medium|
+147|[Insertion Sort List](./0147-insertion-sort-list.js)|Medium|
 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|

solutions/0147-insertion-sort-list.js

@@ -0,0 +1,51 @@
+/**
+ * 147. Insertion Sort List
+ * https://leetcode.com/problems/insertion-sort-list/
+ * Difficulty: Medium
+ *
+ * Given the head of a singly linked list, sort the list using insertion sort, and
+ * return the sorted list's head.
+ *
+ * The steps of the insertion sort algorithm:
+ * - Insertion sort iterates, consuming one input element each repetition and growing
+ *   a sorted output list.
+ * - At each iteration, insertion sort removes one element from the input data, finds
+ *   the location it belongs within the sorted list and inserts it there.
+ * - It repeats until no input elements remain.
+ *
+ * The following is a graphical example of the insertion sort algorithm. The partially
+ * sorted list (black) initially contains only the first element in the list. One
+ * element (red) is removed from the input data and inserted in-place into the sorted
+ * list with each iteration.
+ */
+
+/**
+ * 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 insertionSortList = function(head) {
+  const result = new ListNode(0);
+
+  while (head) {
+    const tail = head;
+    let current = result;
+    head = head.next;
+
+    while (current) {
+      if (!current.next || tail.val <= current.next.val) {
+        [current.next, tail.next] = [tail, current.next];
+        break;
+      }
+      current = current.next;
+    }
+  }
+
+  return result.next;
+};