1480.go

// Source: https://leetcode.com/problems/running-sum-of-1d-array
// Title: Running Sum of 1d Array
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

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