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1588. 所有奇数长度子数组的和

给你一个正整数数组 arr ，请你计算所有可能的奇数长度子数组的和。

子数组 定义为原数组中的一个连续子序列。

请你返回 arr 中 所有奇数长度子数组的和 。

 

示例 1：

输入：arr = [1,4,2,5,3]
输出：58
解释：所有奇数长度子数组和它们的和为：
[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
我们将所有值求和得到 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

示例 2：

输入：arr = [1,2]
输出：3
解释：总共只有 2 个长度为奇数的子数组，[1] 和 [2]。它们的和为 3 。

示例 3：

输入：arr = [10,11,12]
输出：66

提示：

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

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

题目思路

最简单的解法是遍历数组 arr 中的每个长度为奇数的子数组，计算这些子数组的和。由于只需要计算所有长度为奇数的子数组的和，不需要分别计算每个子数组的和，因此只需要维护一个变量sum 存储总和即可。

实现方面，令数组 arr 的长度为 nn，子数组的开始下标为 start，长度为 length，结束下标为 end，则有 0≤start≤end<n，length=end−start+1 为奇数。遍历符合上述条件的子数组，计算所有长度为奇数的子数组的和。

代码

class Solution {
    public int sumOddLengthSubarrays(int[] arr) {
        int sum = 0;
        int n = arr.length;
        for (int start = 0; start < n; start++) {
            for (int length = 1; start + length <= n; length += 2) {
                int end = start + length - 1;
                for (int i = start; i <= end; i++) {
                    sum += arr[i];
                }
            }
        }
        return sum;
    }
}