1480. 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].

Constraints:

• 1 <= nums.length <= 1000
• -10^6 <= nums[i] <= 10^6

时间复杂度：O(n)

空间复杂度：O(1)

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