1588. Sum of All Odd Length Subarrays

QUESTION:

Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.

A subarray is a contiguous subsequence of the array.

Return the sum of all odd-length subarrays of arr.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66

Constraints:

1 <= arr.length <= 100
1 <= arr[i] <= 1000

EXPLANATION:

首先理一下思路, 可以采用例子中的方式, 首先确定长度,或者采用另外一种方式,也就是首先确定位置,长度是可变的.
那么我们就可以: 首先确定需要开始计算的位置,然后每次让长度+2,计算出这段距离的和,再让长度+2, 知道长度超出数组.

SOLUTION:

class Solution {
    fun sumOddLengthSubarrays(arr: IntArray): Int {
        var result = 0;
        for (i in arr.indices) {
            var length = 0;
            while(i+length<=arr.size-1) {
                result+=sum(arr,i,i+length)
                length+=2
            }
        }
        return result
    }

    fun sum(arr: IntArray,start: Int,end:Int) :Int {
        var sum = 0
        for (i in start..end) {
            sum+=arr[i]
        }
        return sum
    }
}
--> return <--