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Java-based LeetCode algorithm problem solutions, regularly updated
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`