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Longest increasing subsequence: Difference between revisions
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</pre>
=={{header|M2000 Interpreter}}==
stack:=stackitem(L(i)), ! stack(L(j)) returns a refence to a new stack object, with the first item on L(i) (which is a reference to stack object) and merge using ! the copy of L(j) stack.
<lang M2000 Interpreter>
Module LIS_example {
Function LIS {
LD=Stack.Size-1
dim L(0 to LD)
For i=0 to LD : Read V: L(i):=Stack:=V:next
M=1
M1=LD
for i=LD-1 to 0
for j=LD to i+1
if stackitem(L(i))<stackitem(L(j)) then
if len(L(i))<=len(L(j)) then L(i) =stack:=stackitem(L(i)), ! stack(L(j))
end if
next
if len(L(i))>=M then M1=i:M=Len(L(i))
next
=L(M1)
}
Const seq$="sequence", subseq$="Longest increasing subsequence"
Document doc$
Disp(seq$, Stack:=3,2,6,4,5,1)
Disp(subseq$, Lis(3,2,6,4,5,1))
Disp(seq$, Stack:=0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15)
Disp(subseq$, LIS(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15))
Print #-2,Doc$
Clipboard Doc$
Sub Disp(title$, m)
local n=each(m), s$
while n
s$+=", "+str$(stackitem(n),"")
end while
s$=trim$(mid$(s$, 2))
Doc$=title$+": "+s$+{
}
End Sub
}
LIS_example
</lang>
{{out}}
<pre style="height:30ex;overflow:scroll">
sequence: 3, 2, 6, 4, 5, 1
Longest increasing subsequence: 3, 4, 5
sequence: 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15
Longest increasing subsequence: 0, 2, 6, 9, 11, 15
</pre >
=={{header|Mathematica}}==
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