leetcode-1480-easy

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

思路一：直接遍历

    public int[] runningSum(int[] nums) {
        int[] result = new int[nums.length];
        
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            result[i] = sum;
        }

        return result;
    }