1480. Running Sum of 1d Array

1. Question

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.

2. Examples

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]

3. Constraints

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

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/running-sum-of-1d-array 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

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