implemented bubblesort in coffeescript and added it to book.json (algorithm-archivists#651)

Author: c252

.editorconfig

@@ -157,3 +157,8 @@ indent_size = 0
 trim_trailing_whitespace = false
 insert_final_newline = false
 end_of_line = lf
+
+# CoffeeScript
+[*.coffee]
+indent_style = space
+indent_size = 2

book.json

@@ -193,6 +193,10 @@
           "lang": "ss",
           "name": "Scheme"
         },
+        {
+          "lang": "coffee",
+          "name": "CoffeeScript"
+        },
         {
           "lang": "kotlin",
           "name": "Kotlin"

contents/bubble_sort/bubble_sort.md

@@ -72,6 +72,8 @@ This means that we need to go through the vector $$\mathcal{O}(n^2)$$ times with
 <p>
   <img  class="center" src="code/scratch/bubble_sort.svg" width="400" />
 </p>
+{% sample lang="coffeescript" %}
+[import:1-6, lang:"coffeescript"](code/coffeescript/bubblesort.coffee)
 {% endmethod %}
 
 ... And that's it for the simplest bubble sort method.
@@ -147,6 +149,8 @@ Trust me, there are plenty of more complicated algorithms that do precisely the
 [import, lang:"bash"](code/bash/bubble_sort.bash)
 {% sample lang="scratch" %}
 The code snippet was taken from this [Scratch project](https://scratch.mit.edu/projects/316483792)
+{% sample lang="coffeescript" %}
+[import, lang:"coffeescript"](code/coffeescript/bubblesort.coffee)
 {% endmethod %}
 
 <script>

contents/bubble_sort/code/coffeescript/bubblesort.coffee

@@ -0,0 +1,13 @@
+bubbleSort = (a) ->
+  n = a.length
+  for i in [0 .. n]
+    for j in [0 .. n - 1]
+      if a[j] > a[j + 1]
+        [a[j + 1], a[j]] = [a[j], a[j + 1]]
+
+main = () ->
+  a = [1, 3, 2, 4, 5, 10, 50, 7, 1.5, 0.3]
+  bubble_sort(a)
+  console.log(a)
+
+main()