LeetCode : Running Sum of 1d Array

yeonju·2020년 8월 3일
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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]

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

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