1588. Sum of All Odd Length Subarrays

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

这题一开始找各个数字出现的规律，后来找了半天没发现规律，估计自己太笨了，直接暴力破解吧。循环里面可以使用滑动窗口的概念来优化子串的求和运算，减去前一个数字，加上最后一个数字即可。

class SumOfAllOddLengthSubarrays : public Solution {
public:
    void Exec() {

    }
    int sumOddLengthSubarrays(vector<int>& arr) {
        int sum = 0;
        for (int i = 1; i <= arr.size(); i += 2) {
            int prevSum = 0;
            for (int j = 0; j < arr.size() - i + 1; j++) {
                if (j == 0) {
                    for (int k = 0; k < i; k++) {
                        prevSum += arr[j + k];
                    }
                } else {
                    prevSum += arr[j + i - 1] - arr[j - 1];
                }
                sum += prevSum;
            }
        }
        return sum;
    }
};