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推廣網站開發,包含 Laravel 和 Kotlin 後端撰寫、自動化測試、讀書心得等。Taiwan Kotlin User Group 管理員。

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Problem 44

Description

This problem was asked by Google.

We can determine how "out of order" an array A is by counting the number of inversions it has. Two elements A[i] and A[j] form an inversion if A[i] > A[j] but i < j. That is, a smaller element appears after a larger element.

Given an array, count the number of inversions it has. Do this faster than O(N^2) time.

You may assume each element in the array is distinct.

For example, a sorted list has zero inversions. The array [2, 4, 1, 3, 5] has three inversions: (2, 1), (4, 1), and (4, 3). The array [5, 4, 3, 2, 1] has ten inversions: every distinct pair forms an inversion.

Solutions