[LeetCode] 1480. Running Sum of 1d Array c++

• 문제

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.

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].

 

• 문제 해결

class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        
        int len = nums.size();
        vector<int> answer(len, 0);
        
        answer[0] = nums[0];
        for(int i = 1; i < len; i++){
            answer[i] = answer[i - 1] + nums[i];
        }
        
        
        return answer;
    }
};

다른 풀이

vector<int> runningSum(vector<int>& nums) {
        int i = 1;
        while (i<nums.size()){
            nums[i]+=nums[i-1];
            i++;
        }
        return nums;
    }