Revisions to Print all lexicographically increasing numbers under 10000

Improved formatting; included link to tio

JavascriptJavaScript, 83 bytes

Borrowed some ideas from @Shaggy, but without recursion.

[...Array(1e4)].reduce((s,_,n)=>s+=([...n+''].every(x=>y<(y=x),y=0))?'\n'+n:'','0')

[...Array(1e4)].reduce((s,_,n)=>s+=([...n+''].every(x=>y<(y=x),y=0))?'\n'+n:'','0')Try it online!

Borrowed some ideas from @Shaggy, but without recursion.

[...Array(1e4)].reduce((s,_,n)=>s+=([...n+''].every(x=>y<(y=x),y=0))?'\n'+n:'','0')

Borrowed some ideas from @Shaggy, but without recursion.

[...Array(1e4)].reduce((s,_,n)=>s+=([...n+''].every(x=>y<(y=x),y=0))?'\n'+n:'','0')

added 67 characters in body

edited Nov 20, 2018 at 20:32

Borrowed some ideas from @Shaggy, but without recursion.

n=0;while(n<1e4){a=++n+"";[[...a]Array(1e4)].sortreduce((s,_,n)=>s+=([...n+''].every(x=>y<(vy=x),iy=0)=>v==a[i])&&console.log(n?'\n'+n:'','0')}

n=0;while(n<1e4){a=++n+"";[...a].sort().every((v,i)=>v==a[i])&&console.log(n)}

Borrowed some ideas from @Shaggy, but without recursion.

[...Array(1e4)].reduce((s,_,n)=>s+=([...n+''].every(x=>y<(y=x),y=0))?'\n'+n:'','0')

answered Nov 20, 2018 at 10:43

n=0;while(n<1e4){a=++n+"";[...a].sort().every((v,i)=>v==a[i])&&console.log(n)}