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

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给定一个数组，求出数组的前缀和。简单题，代码如下：

class Solution {
public:
	vector<int> runningSum(vector<int>& nums) {
		int n = nums.size();
		vector<int> ans(n, 0);
		ans[0] = nums[0];
		for (int i = 1; i < n; ++i)ans[i] = ans[i - 1] + nums[i];
		return ans;
	}
};

本代码提交AC，用时12MS。