1480. Running Sum of 1d Array

Problem Solving

-----------Problem-----------

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: 01

class Solution {
    public int[] runningSum(int[] nums) {     
int[] output = new int[nums.length];

output[0] = nums[0];

for(int i = 1; i<nums.length; i++ ){
 output[i]= nums[i] + output[i - 1] ;
         System.out.println(output[i]);
}
           return output;
    }
}

Solution: 02

class Solution {
  public int[] runningSum(int[] nums) {

for (int i= 1; i < nums.length; i++) {
nums[i] += nums[i - 1];

 System.out.println(nums[i]);
};

return nums;
   }
}