README_EN.md

1480. Running Sum of 1d Array

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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

Python3

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        for i in range(1, len(nums)):
            nums[i] += nums[i - 1]
        return nums

Java

class Solution {
    public int[] runningSum(int[] nums) {
        for (int i = 1; i < nums.length; ++i) {
            nums[i] += nums[i - 1];
        }
        return nums;
    }
}

C++

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

Go

func runningSum(nums []int) []int {
	for i := 1; i < len(nums); i++ {
		nums[i] += nums[i-1]
	}
	return nums
}

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