Running Sum of 1d Array
Problem Statement - link #
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.
Examples #
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
Solutions #
class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
for(int i=1;i<nums.size();i++)
nums[i]+=nums[i-1];
return nums;
}
};