On 06/08/2010 04:53 PM, Simen kjaeraas wrote:
Simen kjaeraas <simen.kja...@gmail.com> wrote:
Andrei Alexandrescu <seewebsiteforem...@erdani.org> wrote
max of n elements is O(n).
T max( T )( T[] values ) {
T result = values[0];
foreach ( i, e; values[1..$] ) {
if ( max( values[i+1..$] ) > result ) {
result = max( values[i+1..$] );
}
}
return result;
}
Better:
T max( T )( T[] values ) {
if ( values.length == 1 ) {
return values[0];
} else {
return max( values[0..values.length/2] ) > max(
values[values.length/2..$] ) ? max( values[0..values.length/2] ) : max(
values[values.length/2..$] );
}
}
I'm not sure I understand the point. Could you please elaborate?
Andrei
