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64 lines
1.1 KiB
Markdown
64 lines
1.1 KiB
Markdown
# [1480. Running Sum of 1d Array](https://leetcode.com/problems/running-sum-of-1d-array/)
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## 题目
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Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums[0]…nums[i])`.
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Return the running sum of `nums`.
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**Example 1**:
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```
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Input: nums = [1,2,3,4]
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Output: [1,3,6,10]
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Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
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```
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**Example 2**:
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```
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Input: nums = [1,1,1,1,1]
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Output: [1,2,3,4,5]
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Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
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```
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**Example 3**:
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```
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Input: nums = [3,1,2,10,1]
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Output: [3,4,6,16,17]
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```
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**Constraints**:
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- `1 <= nums.length <= 1000`
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- `-10^6 <= nums[i] <= 10^6`
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## 题目大意
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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。
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## 解题思路
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- 简单题，按照题意依次循环计算前缀和即可。
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## 代码
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```go
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package leetcode
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func runningSum(nums []int) []int {
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dp := make([]int, len(nums)+1)
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dp[0] = 0
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for i := 1; i <= len(nums); i++ {
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dp[i] = dp[i-1] + nums[i-1]
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}
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return dp[1:]
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}
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``` |