Add solution #338

Author: JoshCrozier

README.md

@@ -169,6 +169,7 @@
 326|[Power of Three](./0326-power-of-three.js)|Easy|
 328|[Odd Even Linked List](./0328-odd-even-linked-list.js)|Medium|
 334|[Increasing Triplet Subsequence](./0334-increasing-triplet-subsequence.js)|Medium|
+338|[Counting Bits](./0338-counting-bits.js)|Easy|
 342|[Power of Four](./0342-power-of-four.js)|Easy|
 344|[Reverse String](./0344-reverse-string.js)|Easy|
 345|[Reverse Vowels of a String](./0345-reverse-vowels-of-a-string.js)|Easy|

solutions/0338-counting-bits.js

@@ -0,0 +1,20 @@
+/**
+ * 338. Counting Bits
+ * https://leetcode.com/problems/counting-bits/
+ * Difficulty: Easy
+ *
+ * Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n),
+ * ans[i] is the number of 1's in the binary representation of i.
+ */
+
+/**
+ * @param {number} n
+ * @return {number[]}
+ */
+var countBits = function(n) {
+  const result = new Array(n + 1).fill(0);
+  for (let i = 0; i <= n; i++) {
+    result[i] = result[i >> 1] + (i & 1);
+  }
+  return result;
+};