Description
https://leetcode.com/problems/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.
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
Explanation
Keep track of the existing sum.
Python Solution
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
results = []
for i, num in enumerate(nums):
if i == 0:
results.append(num)
else:
results.append(results[i - 1] + nums[i])
return results
- Time Complexity: O(N).
- Space Complexity: O(N).