English
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
Solution:
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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