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

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。

• 简单题，按照题意依次循环计算前缀和即可。

package leetcode funcrunningSum(nums []int) []int { dp:=make([]int, len(nums)+1) dp[0] =0fori:=1; i<=len(nums); i++ { dp[i] =dp[i-1] +nums[i-1] } returndp[1:] }