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
}