1480. Running Sum of 1d Array

题目

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

题目大意

给你一个数组 nums 。数组「动态和」的计算公式为：runningSum[i] = sum(nums[0]…nums[i]) 。请返回 nums 的动态和。

解题思路

• 简单题，按照题意依次循环计算前缀和即可。

代码

First Commit

func runningSum(nums []int) []int {
	res := []int{nums[0]}
	for i := 1; i < len(nums); i++ {
		res = append(res, nums[i]+res[i-1])
	}
	return res
}

Patch 1

fix append 扩容

func runningSum(nums []int) []int {
	length := len(nums)
	res := make([]int, length+1)
	res[0] = nums[0]
	for i := 1; i < length; i++ {
		res[i] = res[i-1] + nums[i]
	}
    return res[:len(res)-1]
}

Patch 2

无需创建新变量

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