Question

对两个已排序数组的元素进行排序，而不使用.sort方法，并将已排序的元素放入RUBY中的新数组中

我有两个数组，如下：

a = [1,4,6,9]
b = [2,5,7,8]

我的输出应该是：

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

我该怎么做？

Answer 1

你总是希望利用要合并的两个数组已经排序的事实。这是一种方法：

第一行是避免变异（更改）b或> dput(daata) structure(list(P1 = structure(c(1L, 1L, 3L, 3L, 5L, 5L, 5L, 5L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 2L, 2L), .Label = c("Apple", "Grape", "Orange", "Peach", "Tomato"), class = "factor"), P2 = structure(c(4L, 4L, 3L, 3L, 5L, 5L, 5L, 5L, 6L, 6L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 6L, 6L), .Label = c("Banana", "Cucumber", "Lemon", "Orange", "Potato", "Tomato"), class = "factor"), P1_location_subacon = structure(c(NA, NA, 1L, 1L, 1L, 1L, 1L, 1L, NA, NA, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("Fridge", "Table"), class = "factor"), P1_location_all_predictors = structure(c(2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("Table,Desk,Bag,Fridge,Bed,Shelf,Chair", "Table,Shelf,Cupboard,Bed,Fridge", "Table,Shelf,Fridge"), class = "factor"), P2_location_subacon = structure(c(1L, 1L, 1L, 1L, NA, NA, NA, NA, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("Fridge", "Shelf"), class = "factor"), P2_location_all_predictors = structure(c(3L, 3L, 2L, 2L, 1L, 1L, 1L, 1L, 3L, 3L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("Shelf,Fridge", "Shelf,Fridge,Bed", "Table,Shelf,Fridge"), class = "factor")), .Names = c("P1", "P2", "P1_location_subacon", "P1_location_all_predictors", "P2_location_subacon", "P2_location_all_predictors"), row.names = c(NA, -20L), class = "data.frame")。

Answer 2

这是一个可以帮助你做你想要的功能。

def my_sort_function(a,b)
  n = a.length + b.length
  if a.length < b.length
    min = a.length
  else
    min = b.length
  end
  c=[]

    min.times do |i|
      if a[i] > b[i]
        c << b[i]
        c  << a[i]
      else
        c << a[i]
        c  << b[i]
        end
    end
  if min == a.length

    (c << b[(min)..(b.length-1)]).flatten!

  else
    (c << a[(min)..(a.length-1)]).flatten!

  end

  loop do
    swapped = false

    (n-1).times do |i|
      if c[i] > c[i+1]
        c[i], c[i+1] = c[i+1], c[i]
        swapped = true
      end
    end

    break if not swapped
  end
  c
end

a = [1,4,6]
b = [2,5,7,8]

puts a    #=> a:[1,4,6]
puts b    #=> b:[2,5,7,8]

c= my_sort_function(a,b)
puts c    #=> c:[1,2,4,5,6,7,8]
puts "sorting Done!"

在不使用.sort的情况下在ruby中排序

2 个答案: