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

풀이

var runningSum = function(nums) {
    for(let i = 1; i < nums.length; i++){
        nums[i] = nums[i] + nums[i-1];
    }
    return nums;
};

Reduce를 이용한 방법.

var runningSum = function(nums) {
    nums.reduce((acc, cur, i, arr) => {
        return nums[i] = acc+cur;
    })
    return nums;
};

둘다 Runtime : 84ms, Memory : 39.1MB

MDN-Reduce