1588.所有自然数长度子节点的和(SumofAllOddLengthSubarrays)javascript

题目来自于： https://leetcode-cn.com/problems/sum-of-all-odd-length-subarrays/

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

子数组 定义为原数组中的一个连续子序列。
A subarray is a contiguous subsequence of the array.

请你返回 arr 中 所有奇数长度子数组的和 。
Return the sum of all odd-length subarrays of arr.

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

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

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 。

Example 2:

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

示例 3：

输入：arr = [10,11,12]
输出：66

Example 3:

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

Constraints:

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

// 数组求和

var sumOddLengthSubarrays = function (arr) {
    let sum = 0
    //i,   1，3，5...
    for (var i = 1; i <= arr.length; i += 2) {
        for (var j = 0; j <= arr.length - i; j++) {
            sum += arr.slice(j, j + i).reduce((pre, next) => pre+ next)
        }
    }
    return sum
};

执行用时：88 ms, 在所有 JavaScript 提交中击败了68.97%的用户
内存消耗：43.3 MB, 在所有 JavaScript 提交中击败了23.95%的用户

//数组中元素求和
var arr = [1, 2, 3, 4, 5]
var total = arr.reduce(function (prev, next) {
    return prev + next
}, 0)
console.log(total)

var array1 = [1, , 3, , 5];
var array2 = array1.flat();
// array2: [1, 3, 5]