1480. Running Sum of 1d Array

Question

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.

Solution

类似动态规划，遍历数组，并将当前位置的元素和上一个位置的元素进行加和。

Code

class Solution {
    public int[] runningSum(int[] nums) {
        for(int i = 1; i < nums.length; i++){
            nums[i] += nums[i-1];
        }
        return nums;
    }
}