[ Leetcode ] 1480. Running Sum of 1d Array - Java

 

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

 

 
 class Solution {
    public int[] runningSum(int[] nums) {
        
        int length = nums.length;
        int[] sumNums = new int[length];
        
        for(int i=0; i< length; i++){
            for(int j=i; j< length;j++){
                sumNums[j] += nums[i];
            }
        }
        return sumNums;
    }
 }
 

 

 

여기서 어떻게 더 속도를 빠르게 할지는 고민을 좀 해봐야겠다

++ 100%!

 
 class Solution { 
	public int[] runningSum(int[] nums) {
    
		for(int i=0; i < nums.length-1; i++) {
			nums[i+1] += nums[i];
		}
		return nums;
	}
 }