Leetcode - Running Sum of 1d Array Solution
Poor Man
- Published
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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:
Text
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:
Text
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:
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Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]Constraints:
1 <= nums.length <= 1000-10^6 <= nums[i] <= 10^6
Solution in Python#
Approach 1
Python
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
sum_arr = [nums[0]]
for i in nums[1:]:
sum_arr.append(i+sum_arr[-1])
return sum_arrApproach 2
Python
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
total = 0
sum_arr = []
for i in nums:
total += i
sum_arr.append(total)
return sum_arr