1480. Running Sum of 1d Array (E)

Running Sum of 1d Array (E)

题目

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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题意

求数组中每一个前缀数组的和。

思路

加就完事儿了。

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代码实现

Java

class Solution {
    public int[] runningSum(int[] nums) {
        int[] ans = new int[nums.length];
        int sum = 0;
        
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            ans[i] = sum;
        }
        
        return ans;
    }
}