From: Jason Creighton Date: 2004-07-13T13:07:22+09:00 Subject: Re: A little algorithmic help requested... On Mon, 12 Jul 2004 15:47:41 +0200, "Robert Klemme" wrote: > > "Jason Creighton" schrieb im > Newsbeitrag > news:20040711153514.4b895d41.androflux@softhome.net.remove.to.reply... > > On Sun, 11 Jul 2004 16:10:34 +0900, > > Hal Fulton wrote: > > > > > Here's a problem my tired brain is having trouble with. > > > > > > Given a sorted array of integers, convert them into as many > > > ranges as possible (ranges of three or more). > > > > > > Example: > > > [1,2,3,4,6,7,8,11,12,15,16,17] ==> [1..4,6..8,11,12,15..17] > > > > > > How would *you* do this? > > > > Here's one that doesn't follow that "need at least three of more" > > limitation, but it's how *I* would do it, because it returns an array of > > *only* ranges, which seems like it would be more fun to deal with that a > > mix of ranges and numbers. > > > > module Enumerable > > def to_ranges > > ranges = Array.new > > self.sort.each do |e| > > if ranges[-1] == nil or ranges[-1].end.succ != e > > ranges << Range.new(e,e) > > next > > end > > ranges[-1] = Range.new(ranges[-1].begin, e) > > end > > return ranges > > end > > end > > Nice and short, although I'd use "else" instead of "next"! :-) Oh, right, that would be easier. :) > However, there is a performance drawback: you recreate ranges all over > again. In the worst case of an array that contains all numbers from 1 to > 1000 you create 999 Range instances and keep only one of them. That's not > efficient. I realized I was doing this, but I didn't think the performance hit would be that much. But doing it a nicer was is much faster: ~/prog/rb$ ruby range-problem.rb user system total real to_ranges 2.340000 0.000000 2.340000 ( 2.338682) to_ranges2 0.410000 0.000000 0.410000 ( 0.400007) range-problem.rb: #! /usr/bin/ruby -w # NOTE: to_ranges assumes sorted object. module Enumerable def to_ranges ranges = Array.new self.each do |e| if ranges[-1] == nil or ranges[-1].end.succ != e ranges << Range.new(e,e) else ranges[-1] = Range.new(ranges[-1].begin, e) end end return ranges end def to_ranges2 ranges = Array.new low, high = nil, nil self.each do |e| if low == nil # First item in collection low, high = e,e elsif high.succ != e # We hit something out of sequence ranges << Range.new(low,high) low, high = e, e else high = e end end ranges << Range.new(low,high) return ranges end end if __FILE__ == $0 require 'benchmark' Benchmark.bm(10) do |b| b.report("to_ranges") { (1..2**16).to_ranges } b.report("to_ranges2") { (1..2**16).to_ranges2 } end end Jason Creighton