1480. Running Sum of 1d Array

Easy
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
​

解法

func runningSum(nums []int) []int {
for i := 1; i < len(nums); i++ {
nums[i] += nums[i-1]
}
​
return nums
}