3 ms·

Bubble sort is not robust either

• gus_massa 2y ago
  I refuse to write anything more advanced than Bubble Sort. When I want Nlog(N) I use the sort function in the language. It's strange that they repeat until the list is sorted. I usually repeat the pass N times, ignoring the last items to get a nice exactly N(N-1)/2 time
  • kqr 2y ago
    Why not insertion sort? It's both faster and simpler -- it's what children do when handed playing cards.
    • gus_massa 2y ago
      Stable ordering an (my) lazyness :)
      • kqr 2y ago
        Insertion sort is stable with its normal implementation, for the record.