Skip to content

Latest commit

 

History

History
36 lines (32 loc) · 867 Bytes

1480_running_sum_of_1d_array.md

File metadata and controls

36 lines (32 loc) · 867 Bytes

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:
    def runningSum(self, nums: List[int]) -> List[int]:
        out = []
        runsum = nums[0]
        out.append(runsum)
        for i in range(1, len(nums)):
            out.append(runsum + nums[i])
            runsum += nums[i]

        return out