1480. Running Sum of 1d Array Easy

1/**
2 * [1480] Running Sum of 1d Array
3 *
4 * Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).
5 * 
6 * Return the running sum of nums.
7 * 
8 *  
9 * Example 1:
10 * 
11 * 
12 * Input: nums = [1,2,3,4]
13 * Output: [1,3,6,10]
14 * Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
15 * 
16 * Example 2:
17 * 
18 * 
19 * Input: nums = [1,1,1,1,1]
20 * Output: [1,2,3,4,5]
21 * Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
22 * 
23 * Example 3:
24 * 
25 * 
26 * Input: nums = [3,1,2,10,1]
27 * Output: [3,4,6,16,17]
28 * 
29 * 
30 *  
31 * Constraints:
32 * 
33 * 
34 * 	1 <= nums.length <= 1000
35 * 	-10^6 <= nums[i] <= 10^6
36 * 
37 */
38pub struct Solution {}
39
40// problem: https://leetcode.com/problems/running-sum-of-1d-array/
41// discuss: https://leetcode.com/problems/running-sum-of-1d-array/discuss/?currentPage=1&orderBy=most_votes&query=
42
43// submission codes start here
44
45impl Solution {
46    pub fn running_sum(nums: Vec<i32>) -> Vec<i32> {
47        
48    }
49}
50
51// submission codes end
52
53#[cfg(test)]
54mod tests {
55    use super::*;
56
57    #[test]
58    fn test_1480() {
59    }
60}
61