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Longest increasing subsequence: Difference between revisions

Fixed a comment in D entry
(+ D entry)
(Fixed a comment in D entry)
Line 17:
 
/// Return one of the Longest Increasing Subsequence of
/// items using patience sorting.
T[] lis(T)(T[] items) if (__traits(compiles, T.init < T.init)) {
if (items.empty)
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