Running Sum of 1d Array
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
思路一：直接遍历
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;
}