1480 - Running Sum of 1d Array

1480. Running Sum of 1d Array

![][easy]

Problem Statement:

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
• -10⁶ <= nums[i] <= 10⁶

Solutions:

Java

public class RunningSumOfOneDimensionalArray {
    public static int[] runningSum(int[] nums) {
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            nums[i] = sum;
        }
        return nums;
    }

    public static void main(String[] args) {
        int[] nums = { 1, 1, 1, 1, 1 };
        for (int i : runningSum(nums))
            System.out.print(i + " ");
    }
}

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