Running Sum of 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-----------

Solution: 01

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

output[0] = nums[0];

for(int i = 1; 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;
}
}

The above is the detailed content of Running Sum of Array. For more information, please follow other related articles on the PHP Chinese website!