1480.Running Sum of 1d Array

Tags: Easy Array

Links: https://leetcode.com/problems/running-sum-of-1d-array/

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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:

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

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class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = nums.size();
        vector<int> res(n, 0);
        res[0] = nums[0];
        for (int i = 1; i < n; ++i) {
            res[i] = res[i - 1] + nums[i];
        }

        return res;
    }
};

前缀和的板子。