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1588. 所有奇数长度子数组的和

English Version

题目描述

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

 

进阶：

你可以设计一个 O(n) 时间复杂度的算法解决此问题吗？

解法

“前缀和”实现。

Python3

class Solution:
    def sumOddLengthSubarrays(self, arr: List[int]) -> int:
        n = len(arr)
        presum = [0] * (n + 1)
        for i in range(n):
            presum[i + 1] = presum[i] + arr[i]

        res = 0
        for i in range(n):
            for j in range(0, n, 2):
                if i + j < n:
                    res += presum[i + j + 1] - presum[i]
        return res

Java

class Solution {
    public int sumOddLengthSubarrays(int[] arr) {
        int n = arr.length;
        int[] presum = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            presum[i + 1] = presum[i] + arr[i];
        }
        int res = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; i + j < n; j += 2) {
                res += presum[i + j + 1] - presum[i];
            }
        }
        return res;
    }
}

C++

class Solution {
public:
    int sumOddLengthSubarrays(vector<int>& arr) {
        int n = arr.size();
        int presum[n + 1];
        for (int i = 0; i < n; ++i) presum[i + 1] = presum[i] + arr[i];
        int res = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; i + j < n; j += 2) {
                res += presum[i + j + 1] - presum[i];
            }
        }
        return res;
    }
};

Go

func sumOddLengthSubarrays(arr []int) int {
	n := len(arr)
	presum := make([]int, n+1)
	for i := range arr {
		presum[i+1] = presum[i] + arr[i]
	}
	res := 0
	for i := 0; i < n; i++ {
		for j := 0; i+j < n; j += 2 {
			res += presum[i+j+1] - presum[i]
		}
	}
	return res
}

TypeScript

function sumOddLengthSubarrays(arr: number[]): number {
    const n = arr.length;
    let res = 0;
    for (let i = 1; i <= n; i += 2) {
        let sum = 0;
        for (let j = 0; j < i; j++) {
            sum += arr[j];
        }
        res += sum;
        for (let j = i; j < n; j++) {
            sum -= arr[j - i];
            sum += arr[j];
            res += sum;
        }
    }
    return res;
}

Rust

impl Solution {
    pub fn sum_odd_length_subarrays(arr: Vec<i32>) -> i32 {
        let n = arr.len();
        let mut res = 0;
        let mut i = 1;
        while i <= n {
            let mut sum: i32 = arr[0..i].iter().sum();
            res += sum;
            for j in i..n {
                sum -= arr[j - i];
                sum += arr[j];
                res += sum;
            }
            i += 2;
        }
        res
    }
}

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