Leet Code :Running Sum of 1d Array Solution C++ || Python || Java

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]

Solution :

class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        vector<int>res;
        int ress=0;
        for(int i=0;i<nums.size();i++){
            ress=ress+nums[i];
            res.push_back(ress);
        }
        return res;
    }
};
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