[LeetCode 75/Day 1] 1480. Running Sum of 1d Array

📝 문제 / 1480. Running Sum of 1d Array

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.

Example 1:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

Example 2:

Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

Example 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17] 

Constraints:
1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6

💡 나의 풀이

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var runningSum = function(nums) {
  let sum = 0;
  return nums.map((num) => sum += num);
};