1480. Running Sum of 1d Array.md

1480. Running Sum of 1d Array

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Difficulty : Easy

Related Topics : Array

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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

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Solution

• mine

  • Java

    Runtime: 0 ms, faster than 100.00%, Memory Usage: 41 MB, less than 50.00% of Java online submissions

    // O(N)time O(N)space
    public int[] runningSum(int[] nums) {
        int[] res = new int[nums.length];
        int count = 0;
        for(int i = 0; i < nums.length; i++){
            count+= nums[i];
            res[i] = count;
        }
        return res;
    }