README.md

1480. Running Sum of 1d Array (Easy)

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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/running-sum-of-1d-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        for (int i = 1; i < nums.size(); ++i) nums[i] += nums[i - 1];
        return nums;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/running-sum-of-1d-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        partial_sum(begin(nums), end(nums), begin(nums));
        return nums;
    }
};