[LeetCode] 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

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 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sum-of-all-odd-length-subarrays
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这道题有一个复杂度为 O(n) 的思路，但是如果没做过，面试的时候几乎不会想到。这里我暂时提供一个前缀和的思路。

首先我们用一个数组记录一下 input 数组的前缀和，然后我需要用两层 for 循环，第一层枚举的是子数组的长度，应该只有奇数；第二层枚举的是子数组左端点的下标。

时间O(n^2)

空间O(n)

Java实现

class Solution {
    public int sumOddLengthSubarrays(int[] arr) {
        int[] presum = new int[arr.length + 1];
        for (int i = 0; i < arr.length; i++) {
            presum[i + 1] = presum[i] + arr[i];
        }

        int res = 0;
        // 枚举子数组长度
        for (int i = 1; i <= arr.length; i += 2) {
            // 枚举左端点
            for (int j = 0; j + i - 1 < arr.length; j++) {
                res += presum[j + i] - presum[j];
            }
        }
        return res;
    }
}

LeetCode 题目总结