红宝石阵列 - 查找对角线的总和 [英] Ruby Arrays - Find the sums of the diagonals

问题描述

以前没见过这一个，但我不知道你如何找到在Ruby中二维数组的两条对角线的总和。假设你有一个简单的数组，以3行3列。

 阵列= [1,2,3,4,5,6,7,8,9]
 

我可以通过分解成三个一组

  array.each_sli​​ce（3）.to_a
 

现在是

  [1,2,3]，[4,5,6]，[7,8,9][1,2,3]
[4,5,6]
[7,8,9]
 

在这种情况下，对角线

  1 + 5 + 9 = 15
3 + 5 + 7 = 15
 

所以，总和将 15 + 15 = 30

我想我可以做类似

  diagonal_sum = 0
因为我在0..2
  在0..2Ĵ
    diagonal_sum + =阵列[I] [J]。
  结束
结束
 

解决方案

下面是我的尝试：

 阵列= [1,2,3,4,5,6,7,8,9]
切片= array.each_sli​​ce（3）.to_a
＃由于切片大小为3，我花了2，即3  -  1
（0..2）.MAP {| I |切片[I] [I]}＃=＆GT; [1，5，9]
（0..2）.MAP {| I |切片[Ⅰ] [ -  I-1]}＃=＆GT; [3，5，7]
（0..2）.MAP {| I |切片[I] [I]}。降低+
＃=＆GT; 15
（0..2）.MAP {| I |切片[Ⅰ] [ -  I-1]}。降低+
＃=＆GT; 15
 

根据以上的观察的似乎在一次迭代中，你可以做解决：

  left_diagonal，right_diagoal =（0..2）.each_with_object（[]，[]）做| I，A |
  一个[0]＆下;＆下;切片[I] [I]
  一个[1];＆下;切片[Ⅰ] [ -  I-1]
结束left_diagonal.reduce（+）＃=＆GT; 15
right_diagonal.reduce（+）＃=＆GT; 15
 

添加，的 OOP 的code的风格：

 类方阵
  attr_reader：数组：为了  高清初始化数组，正
    @array = array.each_sli​​ce（N）.to_a
    @order = N
  结束  高清collect_both_diagonal_elements
    （0 ...顺序）.collect_concat {| I | [数组[我] [我]，数组[我] [ -  I-1]}
  结束  高清collect_left_diagonal_elements
    （0 ...顺序）.collect {| I |数组[我] [我]}
  结束  高清collect_right_diagonal_elements
    （0 ...顺序）.collect {| I |阵列[Ⅰ] [ -  I-1]}
  结束  高清sum_of_diagonal_elements类型
    案件类型
    时：所有然后collect_both_diagonal_elements.reduce（0，+）
    当：那么好吧collect_right_diagonal_elements.reduce（0，+）
    时：左，那么collect_left_diagonal_elements.reduce（0，+）
    结束
  结束
结束阵列= [1,2,3,4,5,6,7,8,9]
平方米= SquareMatrix.new阵，3
sqm.collect_both_diagonal_elements＃=＆GT; [1，3，5，5，9日，7]
sqm.sum_of_diagonal_elements：所有＃=＆GT;三十
sqm.collect_left_diagonal_elements＃=＆GT; [1，5，9]
sqm.sum_of_diagonal_elements：左＃=＆GT; 15
sqm.collect_right_diagonal_elements＃=＆GT; [3，5，7]
sqm.sum_of_diagonal_elements：右＃=＆GT; 15
 

Haven't seen this one before, but I was wondering how you can find the sums of both diagonals of a 2D array in Ruby. Say you have a simple array, with 3 rows and 3 columns.

array = [1,2,3,4,5,6,7,8,9]

I can break it into groups of three by using

array.each_slice(3).to_a

Would now be

[1,2,3], [4,5,6], [7,8,9]

[1,2,3]
[4,5,6]
[7,8,9]

In this case, the diagonals are

1 + 5 + 9 = 15
3 + 5 + 7 = 15

So the total sum would be 15 + 15 = 30

I was thinking I could do something like

diagonal_sum = 0
for i in 0..2
  for j in 0..2
    diagonal_sum += array[i][j]
  end
end

解决方案

Here is my try :

array = [1,2,3,4,5,6,7,8,9]
sliced = array.each_slice(3).to_a
# As sliced size is 3, I took 2, i.e. 3 - 1
(0..2).map { |i| sliced[i][i] } #=> [1, 5, 9]
(0..2).map { |i| sliced[i][-i-1] } # => [3, 5, 7]
(0..2).map { |i| sliced[i][i] }.reduce :+
# => 15
(0..2).map { |i| sliced[i][-i-1] }.reduce :+
# => 15

As per the above observation it seems in one iteration you can do solve :

left_diagonal, right_diagoal = (0..2).each_with_object([[], []]) do |i, a|
  a[0] << sliced[i][i]
  a[1] << sliced[i][-i-1]
end

left_diagonal.reduce(:+) # => 15
right_diagonal.reduce(:+) # => 15

Added, OOP style of code :

class SquareMatrix
  attr_reader :array, :order

  def initialize array, n
    @array = array.each_slice(n).to_a
    @order = n
  end

  def collect_both_diagonal_elements
    (0...order).collect_concat { |i| [ array[i][i], array[i][-i-1] ] }
  end

  def collect_left_diagonal_elements
    (0...order).collect { |i| array[i][i] }
  end 

  def collect_right_diagonal_elements
    (0...order).collect { |i| array[i][-i-1] }
  end

  def sum_of_diagonal_elements type
    case type
    when :all   then collect_both_diagonal_elements.reduce(0, :+)
    when :right then collect_right_diagonal_elements.reduce(0, :+)
    when :left  then collect_left_diagonal_elements.reduce(0, :+)
    end
  end
end

array = [1,2,3,4,5,6,7,8,9]
sqm   = SquareMatrix.new array, 3
sqm.collect_both_diagonal_elements # => [1, 3, 5, 5, 9, 7]
sqm.sum_of_diagonal_elements :all # => 30
sqm.collect_left_diagonal_elements # => [1, 5, 9]
sqm.sum_of_diagonal_elements :left # => 15
sqm.collect_right_diagonal_elements # => [3, 5, 7]
sqm.sum_of_diagonal_elements :right # => 15

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