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; } }