1588. Sum of All Odd Length Subarrays

1. Question

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.

2. Examples

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

3. Constraints

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

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/sum-of-all-odd-length-subarrays 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

暴力

class Solution {
    public int sumOddLengthSubarrays(int[] arr) {
        int sum = 0;
        for (int i = 1; i <= arr.length; i+=2) {
            for (int j = 0; j + i <= arr.length; j++) {
                for (int l = 0; l < i; l++) {
                    sum += arr[j + l];
                }
            }
        }
        return sum;
    }
}
class Solution {
    public int sumOddLengthSubarrays(int[] arr) {
        int len = arr.length, res = 0;
        for(int i = 0; i < len; i ++){
            int LeftOdd = (i+1)/2, LeftEven = i/2+1;
            int RightOdd = (len-i)/2, RightEven = (len-1-i)/2+1;
            res += arr[i]*(LeftOdd*RightOdd + LeftEven*RightEven);
        }
        return res;
    }
}
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