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LeetCode Solution: 5453. Running Sum of 1d Array

5453. 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

解析

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class Solution {
public:
vector<int> runningSum(vector<int>& nums)
{
std::vector<int> sum(nums);
for (int i = 1; i < sum.size(); ++i)
sum[i] += sum[i-1];

return sum;
}
};

前缀和求和问题,每次相加为数组之前的元素就行了

时间复杂度O(n)

Author: Jsthcit
Link: http://jsthcitpizifly.com/2020/06/15/LeetCode-Solution-5453/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.
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