Running Sum of 1D Array
Program Description: 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 Solution: class Solution { public int[] runningSum(int[] nums) { int sm = 0; int[] runningSum = new int[nums.length]; for(int i=0; i < nums.length; i++){ sm += nums[i]; runningSum[i] = sm; } ...