Running Sum of 1d Array
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.
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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
C++ Solution
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class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
vector<int> ret;
int last = 0;
for(int i = 0 ; i < nums.size(); i++){
ret.push_back(last + nums[i]);
last = ret.back();
}
return ret;
}
};
