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我已经实现了计算数组反转的算法（它也合并了数组），但为了使它更整洁，我想对输出进行排序和计数。

方法 *count_and_sort* 返回数组 [ number_of_inversions, [sorted_array] ]。方法inversions和 *merge_sort* 应该分别返回 number_of_inversions 和 [sorted_array] ，但我得到NoMethodError

我错过了什么？

class Array

    def inversions
        count_and_sort[0]
    end

    def merge_sort
        count_and_sort[1]
    end

    def merge_sort!
        replace merge_sort
    end

private

    def count_and_sort
        return [0, self] if self.size <= 1
        left, right = partition.with_index{ |value, index| index < self.size / 2}
        left = left.count_and_sort
        right = right.count_and_sort
        merged = merge(left[1], right[1])

        [left[0] + right[0] + merged[0], merged[1]]
    end

    def merge(left, right)
        inversions = 0
        merged = []
        until left.empty? || right.empty?
            if left.first < right.first
                merged << left.shift
            else
                merged << right.shift
                inversions += left.size
            end
        end

        merged += left += right
        [inversions, merged]
    end
end

p [*1..20].shuffle.inversions

我尝试用

def inversions
    result = self.send(:count_and_sort)
    result[0]
end

但结果是一样的。

PS欢迎任何其他批评和评论；）

4

1 回答 1

2

错误在count_and_sort方法内部，如错误消息所述：

... in 'count_and_sort': private method 'count_and_sort' called for ...

错误的行是：

left = left.count_and_sort
right = right.count_and_sort

Ruby 中的私有方法不能用显式接收器调用，它们必须在 上隐式调用self。因此，您可以使该count_and_sort方法受到保护，也可以使用sendonleft和调用它right。在这种情况下，我会建议前者。

于 2013-07-09T09:11:55.027 回答