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| -rw-r--r-- | challenge-131/james-smith/README.md | 10 |
1 files changed, 9 insertions, 1 deletions
diff --git a/challenge-131/james-smith/README.md b/challenge-131/james-smith/README.md index a7987880c8..5c3ce0ff49 100644 --- a/challenge-131/james-smith/README.md +++ b/challenge-131/james-smith/README.md @@ -18,11 +18,19 @@ https://github.com/drbaggy/perlweeklychallenge-club/tree/master/challenge-131/ja ## The solution +There isn't much to the solution, we are going to return the data as an array of arrayrefs each containing consecutive numbers. + + * We start by creating our first arrayref containing the first value. {for the `if` code to work without an edge case we need in element in our first arrayref to compare against) + * We then loop through the values: + * if the next number is 1 greater than the last value in the last arrayref. We push it there, + * otherwise we create a new arrayref and push it on the end of our array. + * We "cheat" a bit with the `if` statement - by replace `if( $a ) { $b } else { $c }` with `($a) ? ($b) : ($c)` this means we can use it inline within a `foreach` loop. + ```perl sub conseq { my @val = @{$_[0]}; my @res = ( [shift @val] ); - ( $_ == 1 + $res[-1][-1] ) ? (push @{$res[-1]},$_) : (push @res,[$_]) foreach @val; + ( $_ == 1 + $res[-1][-1] ) ? (push @{$res[-1]},$_) : (push @res,[$_]) for @val; \@res; } ``` |