1480.running-sum-of-1-d-array.cpp

/*
 * @lc app=leetcode id=1480 lang=cpp
 *
 * [1480] Running Sum of 1d Array
 *
 * https://leetcode.com/problems/running-sum-of-1d-array/description/
 *
 * algorithms
 * Easy (89.68%)
 * Likes:    1278
 * Dislikes: 128
 * Total Accepted:    318.5K
 * Total Submissions: 359.2K
 * Testcase Example:  '[1,2,3,4]'
 *
 * 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
 * 
 */

// @lc code=start
class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        for (int i = 1; i < nums.size(); ++i) {
            nums[i] += nums[i - 1];
        }
        return nums;
    }
};
// @lc code=end