Problem Statement

An O(n2) sorting algorithm which moves elements one at a time into the correct position. The algorithm consists of inserting one element at a time into the previously sorted part of the array, moving higher ranked elements up as necessary. To start off, the first (or smallest, or any arbitrary) element of the unsorted array is considered to be the sorted part. Although insertion sort is an O(n2) algorithm, its simplicity, low overhead, good locality of reference and efficiency make it a good choice in two cases:

The algorithm is as follows (from wikipedia): Writing the algorithm for integers will suffice.

Let's start with the solution:

fun insertionSort = void by List a # sort list in place
  for int i = 1; i < a.length; ++i
    var v = a[i]
    int j
    for j = i - 1; j >= 0 and a[j] > v; --j
      a[j + 1] = a[j]
    end	
    a[j + 1] = v
  end
end
List lists = List[ # a list of lists
  int[4, 65, 2, -31, 0, 99, 83, 782, 1],
  real[5.17, 2, 5.12], 
  text["this", "is", "insertion", "sort"]]
for each List list in lists
  writeLine("Before: " + text!list) # list as text
  insertionSort(list)
  writeLine("After : " + text!list)
  writeLine()
end