Add solution #50

Author: JoshCrozier

README.md

@@ -20,6 +20,7 @@
 36|[Valid Sudoku](./0036-valid-sudoku.js)|Medium|
 41|[First Missing Positive](./0041-first-missing-positive.js)|Hard|
 43|[Multiply Strings](./0043-multiply-strings.js)|Medium|
+50|[Pow(x, n)](./0050-powx-n.js)|Medium|
 54|[Spiral Matrix](./0054-spiral-matrix.js)|Medium|
 58|[Length of Last Word](./0058-length-of-last-word.js)|Easy|
 66|[Plus One](./0066-plus-one.js)|Easy|

solutions/0050-powx-n.js

@@ -0,0 +1,21 @@
+/**
+ * 50. Pow(x, n)
+ * https://leetcode.com/problems/powx-n/
+ * Difficulty: Medium
+ *
+ * Implement `pow(x, n)`, which calculates `x` raised to the power `n` (i.e., `x^n`).
+ */
+
+/**
+ * @param {number} x
+ * @param {number} n
+ * @return {number}
+ */
+var myPow = function(x, n) {
+  if (n === 0) return 1;
+  if (n === 1) return x;
+  if (n < 0) return 1 / myPow(x, -n);
+  if (n % 2 === 0) return myPow(x * x, n / 2);
+
+  return x * myPow(x * x, (n - 1) / 2);
+};