从Ruby中的数组中找到第二个最大的数字。

I have an array a = [3,6,774,24,56,2,64,56,34]. I need to find the second largest number in a single iteration using Ruby. How do I achieve it?

我有一个数组a = [3,6,774,24,56,2,64,56,34]我需要使用Ruby在一次迭代中找到第二大数字。我如何实现它?

5 个解决方案

#1

Simple:

简单:

array.sort[-2]

And you're done :)

你做的:)

#2

sort is probably overkill here, especially for really large arrays. Don't quite understand "single iteration", one line you mean?

这里的sort可能有点过头了，特别是对于真正的大数组。不太理解“单一迭代”，你指的是一条线?

a = [3,6,774,24,56,2,64,56,34]
b = a.shift(2).sort
c = 
  a.inject(b) do |(m2, m), e| 
    case
    when e > m
      [m, e]
    when e > m2
      [e, m]
    else
      [m2, m]
    end
  end
c.first #=> 64

#3

This works, but am not sure for the "single iteration"

这是可行的，但对于“单一迭代”并不确定

a.max(2)[1]

a.max(2)[1]

#4

Just for fun, this method gets the Nth greatest value in a enumerable (you'd use a bisect module to make the insertion into acc more efficient). As pointed out by @Victor, you would only use it when the length of the array is much bigger than n, othersize a simple array.sort[-n] is faster.

仅仅为了好玩，这个方法得到了一个可枚举的第n个最大的值(您将使用一个bisect模块使插入到acc更有效)。正如@Victor指出的，只有当数组的长度比n大得多时，才会使用它，而不是简单数组的大小。[n]更快。

module Enumerable
  def max_nth(n)
    inject([]) do |acc, x|
      (acc + [x]).sort[[acc.size-(n-1), 0].max..-1]
    end.first
  end
end

p [1, 5, 2, 32, 2, 41, 15, 55].max_nth(2) #=> 41

#5

It works too:

它的工作原理:

arr.sort.reverse[1]

#1