查找给定数组中n是否为任意2个数字的和 [英] Find if n exists as a sum of any 2 numbers in the given array

问题描述

我正在尝试查找 n 是否作为所传递数组中任何两个数字的和而存在，如果返回 true 否则为 false ，我的代码存在的问题是 inject 没有按照我的意愿进行迭代。我在做什么错？

I am trying to find whether n exists as a sum of any two numbers in the passed array if so return true else false, the problem with my code is that inject is not iterating as I want it to. What am I doing wrong?

def sum_to_n?(array,n)
  array.each do |i|
    array.inject(i) do |memo,var|
      if memo + var == n
        return true
      else
        return false
      end
    end
  end
end

puts sum_to_n?([1,2,3,4,5],9)

推荐答案

这里是一种方法：

def sum_to_n?(a,n)
  !!a.find{|e| a.include?(n-e)}
end
a = [1,2,3,4,5]
sum_to_n?(a,9) # => true
sum_to_n?(a,11) # => false

如果要获取这两个元素：

If you want to get those 2 elements:

def sum_to_n?(a,n)
  num=a.find{|e| a.include?(n-e)}
  unless num
    puts "not exist"
  else
    p [n-num,num]
  end
end
a = [1,2,3,4,5]
sum_to_n?(a,9)
# >> [5, 4]
sum_to_n?(a,11)
# >> not exist

逻辑

Enumerable＃find 方法每次迭代传递一个数组元素。现在对于任何迭代，例如我有一个元素 e ，然后从n中减去它。现在我只是测试源数组中是否存在（ne），如果找到匹配项 #find 停止查找，并立即返回 e 。如果未找到，则进行下一次迭代。如果 #find 完成其迭代，但未找到（ne），根据文档，它将返回 nil 。

Enumerable#find method passing one array element per iteration.Now for any iteration,say I have an element e,and I subtracted it from n. Now I was just testing that (n-e) is present in the source array.If I found a match #find will stop finding,and immediately will return e.If not found,then it will go for next iteration. If #find completes its iteration,but didn't find (n-e),as per the documentation it will return nil.

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