Question
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].
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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].
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Example 3:
Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]
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My Answer
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
prefix_sum = []
_sum = 0
for i in nums:
_sum = _sum+i
prefix_sum.append(_sum)
return prefix_sum
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Optimized Version
No optimization needed.
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구간합을 만들고 Return만 하면 되는 문제이다.
