LeetCode - Running Sum of 1d Array

문제 설명

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]

Constrains

1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6

Solution

• [첫번째방법] for문으로 풀기
• [두번째방법] Stream-foreach로 풀기.

import java.util.stream.IntStream;

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

    //stream- foreach가 일반 for문 보다 오버헤드는 크다
    public int[] runningSum2(int[] nums) {

        IntStream.range(1,nums.length)
                .forEach(i->{
                    nums[i]=nums[i-1]+nums[i];
                });
        return nums;
    }
}