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Longest increasing subsequence: Difference between revisions

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=={{header|M2000 Interpreter}}==
===Using Stack objects in an array===
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.
 
Line 1,297 ⟶ 1,298:
while n
s$+=", "+str$(stackitem(n),"")
end while
s$=trim$(mid$(s$, 2))
Doc$=title$+": "+s$+{
}
End Sub
}
LIS_example
</lang>
 
<lang M2000 Interpreter>
===Using arrays in an array===
Module LIS_example {
Function LIS {
LD=Stack.Size-1
dim L(0 to LD)
For i=0 to LD : Read V: L(i):=(V,):next
M=1
M1=LD
for i=LD-1 to 0
for j=LD to i+1
if Array(L(i))<Array(L(j)) then
' you can use either is the same
' if len(L(i))<=len(L(j)) then L(i)=Cons((Array(L(i)),), L(j))
if len(L(i))<=len(L(j)) then L(i)=(Array(L(i)),): Append L(i), 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$, (3,2,6,4,5,1))
Disp(subseq$, LIS(3,2,6,4,5,1))
Disp(seq$, (0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15))
Disp(subseq$, 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$+=", "+str$(Array(n),"")
end while
s$=trim$(mid$(s$, 2))
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