Datasets:

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karths

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binary-10IQR-comp

like 0

ERROR: type should be string, got "https://leetcode.com/problems/sum-of-all-odd-length-subarrays/\r\n\r\n```C#\r\nusing System.Linq;\r\n\r\npublic class Solution\r\n{\r\n public int SumOddLengthSubarrays(int[] arr)\r\n {\r\n // 「輸入的數字」（複數）\r\n var inputNumbers = arr;\r\n\r\n // This is how we will solve this in O(n) time.\r\n //\r\n // Say we have inputNumbers like this:\r\n //\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n //\r\n // ---\r\n //\r\n // These are all of the 1-length subarrays:\r\n //\r\n // {n₁}\r\n // {n₂}\r\n // {n₃}\r\n // {n₄}\r\n // {n₅}\r\n // {n₆}\r\n // ...\r\n // {n₋₆}\r\n // {n₋₅}\r\n // {n₋₄}\r\n // {n₋₃}\r\n // {n₋₂}\r\n // {n₋₁}\r\n //\r\n // They add up like this:\r\n //\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n //\r\n // ---\r\n //\r\n // These are all of the 3-length subarrays:\r\n //\r\n // {n₁, n₂, n₃}\r\n // {n₂, n₃, n₄}\r\n // {n₃, n₄, n₅}\r\n // {n₄, n₅, n₆}\r\n // {n₅, n₆, ...\r\n // {n₆, ...\r\n // ...\r\n // ... , n₋₆}\r\n // ... , n₋₆, n₋₅}\r\n // {n₋₆, n₋₅, n₋₄}\r\n // {n₋₅, n₋₄, n₋₃}\r\n // {n₋₄, n₋₃, n₋₂}\r\n // {n₋₃, n₋₂, n₋₁}\r\n //\r\n // They add up like this:\r\n //\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // { n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, }\r\n // { n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, }\r\n //\r\n // It is equivlent to:\r\n //\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // -n₁ -n₋₁\r\n // -n₁ -n₂ -n₋₂ -n₋₁\r\n //\r\n // ---\r\n //\r\n // We can list all 5-length subarrays and add them up. It will\r\n // turn out like this:\r\n //\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // -n₁ -n₋₁\r\n // -n₁ -n₂ -n₋₂ -n₋₁\r\n // -n₁ -n₂ -n₃ -n₋₃ -n₋₂ -n₋₁\r\n // -n₁ -n₂ -n₃ -n₄ -n₋₄ -n₋₃ -n₋₂ -n₋₁\r\n //\r\n // ---\r\n //\r\n // This is the case for 7-length subarrays.\r\n //\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // {n₁, n₂, n₃, n₄, n₅, n₆, ... , n₋₆, n₋₅, n₋₄, n₋₃, n₋₂, n₋₁}\r\n // -n₁ -n₋₁\r\n // -n₁ -n₂ -n₋₂ -n₋₁\r\n // -n₁ -n₂ -n₃ -n₋₃ -n₋₂ -n₋₁\r\n // -n₁ -n₂ -n₃ -n₄ -n₋₄ -n₋₃ -n₋₂ -n₋₁\r\n // -n₁ -n₂ -n₃ -n₄ -n₅ -n₋₅ -n₋₄ -n₋₃ -n₋₂ -n₋₁\r\n // -n₁ -n₂ -n₃ -n₄ -n₅ -n₆ -n₋₆ -n₋₅ -n₋₄ -n₋₃ -n₋₂ -n₋₁\r\n //\r\n // ---\r\n //\r\n // This pattern will continue to hold for all (2m+1)-length\r\n // subarrays, and we will take advantage of it.\r\n\r\n // 「最大的 Subarray 長度」\r\n var maxSubarrayLength = inputNumbers.Length % 2 == 0 ? inputNumbers.Length - 1 : inputNumbers.Length;\r\n\r\n // 「輸出值」\r\n var output = inputNumbers.Sum() * (1 + maxSubarrayLength) * ((maxSubarrayLength - 1) / 2 + 1) / 2;\r\n\r\n // 「『多餘的項數』的數量變數１」\r\n var excessiveTermCount1 = 1;\r\n\r\n // 「『多餘的項數』的數量變數２」\r\n var excessiveTermCount2 = 2;\r\n\r\n // 「『多餘的項數』的數量的差１」\r\n var excessiveTermCountDifference1 = 1;\r\n\r\n // 「『多餘的項數』的數量的差２」\r\n var excessiveTermCountDifference2 = 2;\r\n\r\n for (int i = maxSubarrayLength - 2; i > 0; i -= 2)\r\n {\r\n output -= inputNumbers[i - 1] * excessiveTermCount2;\r\n output -= inputNumbers[i] * excessiveTermCount1;\r\n output -= inputNumbers[inputNumbers.Length - i - 1] * excessiveTermCount1;\r\n output -= inputNumbers[inputNumbers.Length - i] * excessiveTermCount2;\r\n\r\n excessiveTermCountDifference1 += 2;\r\n excessiveTermCountDifference2 += 2;\r\n excessiveTermCount1 += excessiveTermCountDifference1;\r\n excessiveTermCount2 += excessiveTermCountDifference2;\r\n }\r\n\r\n return output;\r\n }\r\n}\r\n```\r\n\r\n# 參考資料\r\n\r\n## 等差數列\r\n\r\nhttps://en.wikipedia.org/wiki/Arithmetic_progression\r\n\r\n* 項數： term\r\n* 差： difference\r\n\r\n---\r\n請參考「刷 LeetCode 練習命名」 https://github.com/EngTW/English-for-Programmers/issues/69 😊"