1480. Running Sum of 1d Array
Description
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
Solutions
Solution 1: Prefix Sum
We directly traverse the array. For the current element $nums[i]$, we add it with the prefix sum $nums[i-1]$ to get the prefix sum $nums[i]$ of the current element.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
1 2 3 |
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1 2 3 4 5 6 7 8 |
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1 2 3 4 5 6 7 |
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1 2 3 4 5 6 |
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1 2 3 4 5 6 |
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1 2 3 4 5 6 7 8 |
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1 2 3 4 5 6 7 8 9 10 11 12 |
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