Add solution #1480

Author: JoshCrozier

README.md

@@ -169,6 +169,7 @@
 1470|[Shuffle the Array](./1470-shuffle-the-array.js)|Easy|
 1472|[Design Browser History](./1472-design-browser-history.js)|Medium|
 1475|[Final Prices With a Special Discount in a Shop](./1475-final-prices-with-a-special-discount-in-a-shop.js)|Easy|
+1480|[Running Sum of 1d Array](./1480-running-sum-of-1d-array.js)|Easy|
 1486|[XOR Operation in an Array](./1486-xor-operation-in-an-array.js)|Easy|
 1598|[Crawler Log Folder](./1598-crawler-log-folder.js)|Easy|
 1780|[Check if Number is a Sum of Powers of Three](./1780-check-if-number-is-a-sum-of-powers-of-three.js)|Medium|

solutions/1480-running-sum-of-1d-array.js

@@ -0,0 +1,21 @@
+/**
+ * 1480. Running Sum of 1d Array
+ * https://leetcode.com/problems/running-sum-of-1d-array/
+ * Difficulty: Easy
+ *
+ * Given an array nums. We define a running sum of an array as
+ * runningSum[i] = sum(nums[0]…nums[i]).
+ *
+ * Return the running sum of nums.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var runningSum = function(nums) {
+  for (let i = 1; i < nums.length; i++) {
+    nums[i] = nums[i - 1] + nums[i];
+  }
+  return nums;
+};