红宝石阵列

本文介绍了红宝石阵列 - 查找对角线的总和的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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 = 153 + 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; 15right_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，3sqm.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; 15sqm.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