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7/27/2019 Ngn xp v biu thc trung t

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Ngn xp v biu thc trung t, hu t, tin t

byhosytanon 13/5/2013, 15:11

1. Khi nim

Mt php ton hai ngi trn tp hp X l mt nh x f: XX X cho (a,b) f(a,b)

thuc A. nh x f khi thng c k hiu bi *, c gi l ton t, cc phn t

a, b c gi l cc hng t (cn gi l ton hng).

Khi vit biu thc biu din php ton ta c th t k hiu ton t trc (k

php tin t), sau (k php hu t) hoc gia (k php trung t) cc ton hng.

Thng thng trong cc biu thc i s v s hc, ta vit k hiu php ton gia

hai hng t, l k php trung t. V d: a + b, a * b,... Khi mt biu thc c nhiu

php ton, ta dng cc cp du ngoc "(", ")" v th t u tin cc php ton ch

r th t thc hin cc php ton. (Cc php ton u quy v php ton 2 ngi.)

Ta cng c th vit hai hng t trc v k hiu ton t sau. Chng hn:

a + b vit l a b +, a * b vit l a b *

Cng c th vit ton t trc, hai ton hng sau. Chng hn:

a + b vit l + a b, a * b vit l * a b

V l thuyt, k php tin t v k php hu t cn c th c m rng cho cc

php ton ba ngi hoc nhiu hn m vn khng phi dng ti du ngoc th

hin u tin cc php ton, tng t vi hm s a bin, cn k php trung t th

khng th. Tuy nhin, trong thc t khng c nhiu php ton a ngi v k php

trung t vn c dng rng ri v thi quen. V d: + a b c c th c hiu l tng

ca 3 s a, b v c trong k php tin t. Tng t, f a b c c th c hiu l hm fca 3 bin a, b v c trong k php tin t.

Tm li:

- Tin t: Vit ton t trc ton hng

- Hu t: Vit ton t sau ton hng

- Trung t: Vit ton t gia hai ton hng.

Ch : Vi cch biu din dng trung t, ta c cc trng hp sau:

- Trung t y du ngoc: Tt c ton t v ton hng u dng 2 ngi v t

trong mt cp du ngoc.

- Trung t khng y du ngoc: Ch c mt s cp ton hng v ton t l cdu ngoc.

- Trung t khng c du ngoc: Biu thc khng cha bt k du ngoc no.

tnh biu thc ca hai dng trung t sau ta cn xt n tnh u tin ca ton t.

Nguyn tc u tin l *, /, % u tin hn + v -.Chuyn i biu thc Trung t y du ngoc sang Hu t
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a. Bi ton

Chuyn biu thc M = M1M2M3...Mn t dng trung t y du ngoc sang hu t

(N).

b. Nhn xt- V M dng trung t y du ngoc nn chc chn mi ton t hai ngi u

nm trong mt cp du ngoc.

- Ta s phi thc hin php ton khi gp du ngoc ng u tin.

c. tng

Duyt t M t tri qua phi:

- Nu gp ton hng th a ton hng vo biu thc ch N.

- Nu gp ton t th a vo mt ngn xp (stack) S no khi to t trc.

- Gp du ngoc m ("(") th b qua, gp du ngoc ng (")") th ly ton t tnh ca S a vo biu thc ch N.

d. Gii thut

CODE:

Input: M: Bi u th c trung t y d u ngo c.

1. Khi to stack S = {}; // Stack S rng.

2. Khi to biu thc ch N = ""; // N cha c phn t no.

3. for i:=1 to |M| do

a. if (Mi l Ton hng) then N:=N+Mi; // a Mi vo N

b. if (Mi l Ton t) then PUSH(Mi,S); // a Mi vo ngn xp S

c. if (Mi l ")" ) then

POP(S,a); // Ly gi tr trn nh ca S a vo a

N:=N+a; // Thm a vo N

Output: N: Biu thc hu t.

e. V d

CODE:

Input: M = (10-(3*((14-2)/(2+4)))) sang hu t ( y M c s

phn t l n=21)

- Khi to: S={}; N="";

- M1 = "(", b qua


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- M2 = "10", S={}; N="10";

- M3 = "-", S={-}; N="10";

- M4 = "(", S={-}; N="10";

- M5 = "3", S={-}; N="10, 3";

- M6 = "*", S={-, *}; N="10, 3";- M7 = "(", S={-, *}; N="10, 3";

- M8 = "(", S={-, *}; N="10, 3";

- M9 = "14", S={-, *}; N="10, 3, 14";

- M10 = "-", S={-, *, -}; N="10, 3, 14";

- M11 = "2", S={-, *, -}; N="10, 3, 14, 2";

- M12 = ")", S={-, *}; N="10, 3, 14, 2, -";

- M13 = "/", S={-, *, /}; N="10, 3, 14, 2, -";

- M14 = "(", S={-, *, /}; N="10, 3, 14, 2, -";

- M15 = "2", S={-, *, /}; N="10, 3, 14, 2, -, 2";- M16 = "+", S={-, *, /, +}; N="10, 3, 14, 2, -, 2";

- M17 = "4", S={-, *, /, +}; N="10, 3, 14, 2, -, 2, 4";

- M18 = ")", S={-, *, /}; N="10, 3, 14, 2, -, 2, 4, +";

- M19 = ")", S={-, *}; N="10, 3, 14, 2, -, 2, 4,

+, /";

- M20 = ")", S={-}; N="10, 3, 14, 2, -, 2, 4,

+, /, *";

- M21 = ")", S={}; N="10, 3, 14, 2, -, 2, 4, +, /,

*, -";Output: N = 10 3 14 2 - 2 4 + / * -

Chuyn i biu thc Trung t khng y du ngoc sang Hut

a. Bi ton

Chuyn biu thc M = M1M2M3...Mn t dng trung t khng y du ngoc sang

hu t (N).

b. Nhn xt

- V M dng trung t khng y du ngoc nn khi gp ton t ta cn xt n
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u tin ca ton t.

c. tng


- Nu gp ton hng th a ton hng vo biu thc ch N.- Nu gp du ngoc m ("(") th a vo mt ngn xp (stack) S no khi to

t trc.

- Nu gp du ngoc ng (")") th ln lt ly cc ton t t nh ca S a vo

biu thc ch N cho n khi gp du ngoc ng (")") u tin, loi b lun du

ngoc ng ra khi stack.

- Nu gp ton t c u tin cao hn ton t hin c trn nh stack th a n

vo stack.

- Nu gp ton t c u tin thp hn hoc bng ton t hin ti trn inh stack

th ly ton t trn nh stack ra a vo biu thc ch N, ng thi a ton t vo nh stack.

- Khi xt ht cc phn t ca biu thc M, ln lt ly cc ton t c trong stack ra

biu thc ch N.

d. Gii thut

CODE:

Input: M: Biu thc trung t khng y du ngoc.

int UT(x){ // x l mt phn t ca M

if ((x="*") OR (x="/") OR (x="%")) ruturn 2;

if ((x="+") OR (x="-")) ruturn 1;

}


2. Khi to biu thc ch N = ""; // N cha c phn t no.

3. for i:=1 to |M| do

a. if (Mi l Ton hng) N:=N+Mi; // a Mi vo N

b. if (Mi l Ton t)

if UT(Top(S)>=UT(Mi){

Pop(S,x);

N:=N+x;

Push(Mi,S);


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}

c. if (Mi l ")" ){

x="";

while(x"("){

N:=N+x; // Thm x vo NPOP(S,x); // Ly gi tr trn nh ca S a vo x

}

}

4. while !IsEmpty(S){

Pop(S,x);

if x"(" N:=N+x;

}


e. V d

CODE:

Input: M = 10-3*(14-2)/(2+4) sang hu t ( y M c s phn t

l n=15)

- Khi to: S={}; N="";

- M1 = "10", S={}; N="10";

- M2 = "-", S={-}; N="10";

- M3 = "3", S={-}; N="10, 3";

- M4 = "*", S={-, *}; N="10, 3";

- M5 = "(", S={-, *, (}; N="10, 3";

- M6 = "14", S={-, *, (}; N="10, 3, 14";

- M7 = "-", S={-, *, (, -}; N="10, 3, 14";

- M8 = "2", S={-, *, (, -}; N="10, 3, 14, 2";

- M9 = ")", S={-, *}; N="10, 3, 14, 2, -";

- M10 = "/", S={-, /}; N="10, 3, 14, 2, -, *";

- M11 = "(", S={-, /, (}; N="10, 3, 14, 2, -, *";

- M12 = "2", S={-, /, (}; N="10, 3, 14, 2, -, *, 2";

- M13 = "+", S={-, /, (, +}; N="10, 3, 14, 2, -, *, 2";

- M14 = "4", S={-, /, (, +}; N="10, 3, 14, 2, -, *, 2,

4";

- M15 = ")", S={-, /}; N="10, 3, 14, 2, -, *, 2,


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4, +";

- M16 = "", S={}; N="10, 3, 14, 2, -, *, 2, 4,

+, /, -";

Output: N = 10 3 14 2 - * 2 4 + / -

Chuyn i biu thc Trung t khng c du ngoc sang Hu t

a. Bi ton

Chuyn biu thc M = M1M2M3...Mn t dng trung t khng c du ngoc sang hut (N).

b. Nhn xt

- y l trng hp ring ca trng hp M khng y du ngoc (trong qu trnh

xt khng gp du ngoc no), ta vn cn xt n u tin ca ton t.

c. tng


- Nu gp ton hng th a ton hng vo biu thc ch N.- Nu gp ton t c u tin cao hn ton t hin c trn nh stack th a n

vo stack.

- Nu gp ton t c u tin thp hn hoc bng ton t hin ti trn inh stack

th ly ton t trn nh stack ra a vo biu thc ch N, ng thi a ton t

vo nh stack.

- Khi xt ht cc phn t ca biu thc M, ln lt ly cc ton t c trong stack ra

biu thc ch N.

d. Gii thut

CODE:

Input: M: Biu thc trung t c du ngoc.


if ((x="*") OR (x="/") OR (x="%")) ruturn 2;
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if ((x="+") OR (x="-")) ruturn 1;

}


2. Khi to biu thc ch N = ""; // N cha c phn t no.3. for i:=1 to |M| do

a. if (Mi l Ton hng) N:=N+Mi; // a Mi vo N

b. if (Mi l Ton t)

if UT(Top(S)>=UT(Mi){

Pop(S,x);

N:=N+x;

Push(Mi,S);

}

4. while !IsEmpty(S){Pop(S,x);

N:=N+x;

}


e. V d

CODE:

Input: M = 10-3*14-2/2+4 sang hu t ( y M c s phn t l

n=11)

- Khi to: S={}; N="";

- M1 = "10", S={}; N="10";

- M2 = "-", S={-}; N="10";

- M3 = "3", S={-}; N="10, 3";

- M4 = "*", S={-, *}; N="10, 3";

- M5 = "14", S={-, *}; N="10, 3, 14";

- M6 = "-", S={-, -}; N="10, 3, 14, *";

- M7 = "2", S={-, -}; N="10, 3, 14, *, 2";

- M8 = "/", S={-, -, /}; N="10, 3, 14, *, 2";

- M9 = "2", S={-, -, /}; N="10, 3, 14, *, 2, 2";

- M10 = "+", S={-, -, +}; N="10, 3, 14, *, 2, 2, /";

- M11 = "4", S={-, -, +}; N="10, 3, 14, *, 2, 2, /,


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4";

- M12 = "", S={}; N="10, 3, 14, *, 2, 2, /, 4,

+, -, -";

Output: N = 10 3 14 * 2 2 / 4 + - -

Tnh gi tr biu thc dng Trung t y du ngoc

a. Bi ton

Tnh gi tr biu thc M = M1M2M3...Mn dng trung t y du ngoc.

b. Nhn xt

- Ta nn s dng mt stack lu tr cc kt qu tnh ton trung gian.

c. tng

Duyt t M t tri qua phi:- Nu gp du ngoc m ("(") th b qua.

- Nu gp ton hng hoc ton t th a chng vo stack.

- Nu gp du ngoc ng (")") th ly 3 phn t trn nh stack (gm 2 ton hng

v mt ton t) ri thc hin php ton, kt qu li ct vo nh stack.

- Khi xt ht M, gi tr cui cng trn nh stack chnh l kt qu tnh ton biu thc

M.

d. Gii thut

CODE:

Input: M: Biu thc trung t y du ngoc.


2. for (i=1;i


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Push((bxa),S); // Tnh gi tr bxa, kt qu li a vo S.

}

}

3. Pop(S,Result);

Output: Result.

e. V d

CODE:

Cu I, thi tuyn sinh cao hc nm 2013

Input: M = (10-(3*((14-2)/(2+4))))

- Khi to: S={};

- M1="(" B qua S={}

- M2="10" a 10 vo S S={10}

- M3="-" a - vo S S={10, -}

- M4="(" B qua S={10, -}

- M5="3" a 3 vo S S={10, -, 3}

- M6="*" a * vo S S={10, -, 3, *}

- M6="(" B qua S={10, -, 3, *}

- M7="(" B qua S={10, -, 3, *}

- M8="14" a 14 vo S S={10, -, 3, *, 14}

- M9="-" a - vo S S={10, -, 3, *, 14,

-}

- M10="2" a 2 vo S S={10, -, 3, *,

14, -, 2}

- M11=")" Tnh 14-2=12 ri vo S S={10, -, 3, *,

12}

- M12="/" a / vo S S={10, -, 3, *,

12, /}

- M13="(" B qua S={10, -, 3, *,

12, /}

- M14="2" a 2 vo S S={10, -, 3, *,

12, /, 2}

- M15="+" a + vo S S={10, -, 3, *,

12, /, 2, +}

- M16="4" a 4 vo S S={10, -, 3, *,


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12, /, 2, +, 4}

- M17=")" Tnh 2+4=6 ri vo S S={10, -, 3, *,

12, /, 6}

- M18=")" Tnh 12/6=2 ri vo S S={10, -, 3, *, 2}

- M19=")" Tnh 3*2=6 ri vo S S={10, -, 6}- M20=")" Tnh 10-6=4 ri vo S S={4}

- M21="" Kt thc vic tnh ton S={4}

Output: Result=4

Tnh gi tr biu thc dng Trung t khng y du ngoc

a. Bi ton

Tnh gi tr biu thc M = M1M2M3...Mn dng trung t khng y du ngoc.

b. Nhn xt- Ta nn s dng hai stack lu tr cc kt qu tnh ton trung gian. Stack Sh lu

tr cc ton hng, stack St lu tr cc ton t v du ngoc m.

- V biu thc M dng Trung t khng y du ngoc nn ta cn xt n u

tin ca cc ton t.

c. tng

Gi s M c cho dng ng (ngha l du ngoc phi i vi nhau tng i mt,

cc phn t ch gm ton t, ton hng hoc du ngoc).

Khi to 2 stack: Sh v St.Duyt t M t tri qua phi:

- Nu gp du ngoc m ("(") th a vo St.

- Nu gp ton hng th a vo Sh.

- Nu gp ton t c u tin cao hn ton t hin c trn nh St th a n vo

St.

- Nu gp ton t c u tin thp hn hoc bng ton t hin c trn nh St th

ly ton t trn nh St cng hai ton hng trn nh Sh, thc hin php ton ri ct

kt qu vo Sh.

- Nu gp du ngoc ng (")") th ly ton t trn nh St cng 2 ton hng trnnh Sh, thc hin php ton, kt qu ct vo Sh. C thc hin nh vy cho n khi

gp du ngoc m u tin trn St, loi b lun du ngoc m ny ra khi St.

- Khi xt ht M, ln lt ly mt ton t trn St v 2 ton hng trn Sh, thc hin

php ton, kt qu ct vo Sh cho n khi Sh ch cn mt phn t (hay n khi St

rng).
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d. Gii thut

CODE:

Input: M: Biu thc trung t khng y du ngoc.


if x thuc {*, /, %} return 2;

if x thuc {+, -} return 1;

if x = "(" return 0

}

1. Khi to stack Sh = St = {}; // Stack Sh, St rng.

2. for (i=1;i


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Pop(Sh,a); // a l mt ton hng ly t Sh.

Pop(St,x); // x l mt ton t ly t St.

Pop(Sh,b); // b l mt ton hng ly t Sh.

Push((bxa),Sh); // Tnh gi tr bxa, kt qu li a vo Sh.

}4. Pop(Sh,Result); // Ly phn t cui cng ca Sh ra kt qu.

Output: Result.

e. V d

CODE:

Input: M = 10-3*(14-2)/(2+4)

- Khi to: Sh={};St={};

- M1="10" a 10 vo Sh Sh={10}

St={}

- M2="-" a - vo St Sh={10}

St={-}

- M3="3" a 3 vo Sh Sh={10, 3}

St={-}

- M4="*" a * vo St Sh={10, 3}

St={-, *}

- M5="(" a ( vo St Sh={10, 3}

St={-, *, (}

- M6="14" a 14 vo Sh Sh={10, 3,

14} St={-, *, (}

- M7="-" a - vo St Sh={10, 3,

14} St={-, *, (, -}

- M8="2" a 2 vo Sh Sh={10, 3,

14, 2} St={-, *, (, -}

- M9=")" Tnh 14-2=12 ri vo Sh, loi b lun ( St

Sh={10, 3, 12} St={-, *}

- M10="/" Tnh 3*12=36 ri a vo Sh, a / vo St

Sh={10, 36} St={-, /}

- M11="(" a ( vo St Sh={10, 36}

St={-, /, (}

- M12="2" a 2 vo Sh Sh={10, 36,


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2} St={-, /, (}

- M13="+" a + vo St Sh={10, 36,

2} St={-, /, (, +}

- M14="4" a 4 vo Sh Sh={10, 36,

2, 4} St={-, /, (, +}- M15=")" Tnh 2+4=6 ri vo Sh, loi b lun ( St

Sh={10, 36, 6} St={-, /}

- Top(St)="/" Tnh 36/6=6 ri a vo Sh

Sh={10, 6} St={-}

- Top(St)="-" Tnh 10-6=4 ri a vo Sh

Sh={4} St={}

- Top(St)="" Dng

Output: Result=Top(Sh) = 4

Tnh gi tr biu thc dng Trung t khng c du ngocbyhosytanon 14/5/2013, 11:13

a. Bi ton

Tnh gi tr biu thc M = M1M2M3...Mn dng trung t khng c du ngoc.

b. Nhn xt

- C th s dng lun th tc tnh ton biu thc dng Trung t khng y du

ngoc thc hin v y ch l trng hp c bit ca trng hp khng y du ngoc.

- Ta vn nn s dng hai stack lu tr cc kt qu tnh ton trung gian. Stack Sh

lu tr cc ton hng, stack St lu tr cc ton t.

- V biu thc M dng Trung t c khng du ngoc nn ta cn xt n u tin

ca cc ton t.

c. tng

Gi s M c cho dng ng (ngha l cc phn t ch gm ton t v ton

hng ).Khi to 2 stack: Sh v St.


- Nu gp ton hng th a vo Sh.

- Nu gp ton t c u tin cao hn ton t hin c trn nh St th a n vo

St.

- Nu gp ton t c u tin thp hn hoc bng ton t hin c trn nh St th
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ly ton t trn nh St cng hai ton hng trn nh Sh, thc hin php ton ri ct

kt qu vo Sh. a ton t ang xt vo St.

- Khi xt ht M, ln lt ly mt ton t trn St v 2 ton hng trn Sh, thc hin

php ton, kt qu ct vo Sh cho n khi Sh ch cn mt phn t (hay n khi St

rng).

d. Gii thut

CODE:

Input: M: Biu thc trung t khng c du ngoc.


if x thuc {*, /, %} return 2;

if x thuc {+, -} return 1;

}

1. Khi to stack Sh = St = {}; // Stack Sh, St rng.

2. for (i=1;i


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e. V d

CODE:

Input: M = 10-3*14-2/2+4

- Khi to: Sh={};St={};

- M1="10" a 10 vo Sh Sh={10}

St={}

- M2="-" a - vo St Sh={10}

St={-}

- M3="3" a 3 vo Sh Sh={10, 3}

St={-}

- M4="*" a * vo St Sh={10, 3}

St={-, *}

- M6="14" a 14 vo Sh Sh={10, 3,

14} St={-, *}

- M7="-" Ly 14*3=42 a vo Sh, a - vo St

Sh={10, 42} St={-, -}

- M8="2" a 2 vo Sh Sh={10, 42,

2} St={-, -}

- M10="/" Da / vo St Sh={10, 42,

2} St={-, -, /}

- M12="2" a 2 vo Sh Sh={10, 42,

2, 2} St={-, -, /}

- M13="+" Ly 2/2=1 a vo Sh, a + vo St

Sh={10, 42, 4} St={-, -, +}

- M14="4" a 4 vo Sh Sh={10, 42,

4, 4} St={-, -, +}

- Top(St)="+" Tnh 4+4=8 ri a vo Sh

Sh={10, 42, 8} St={-, -}

- Top(St)="-" Tnh 42-8=34 ri a vo Sh

Sh={10, 34} St={-}

- Top(St)="-" Tnh 10-34=-24 ri a vo Sh Sh={-

24} St={}

- Top(St)="" Dng


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Output: Result=Top(Sh) = -24

Tnh gi tr biu thc dng Hu tbyhosytanon 14/5/2013, 13:26

a. Bi tonTnh gi tr biu thc M = M1M2M3...Mn dng hu t.

b. Nhn xt

- Do M dng hu t nn ta khng cn quan tm n u tin ca cc ton t.

- Ta vn dng 1 ngn xp lu kt qu trnh ton trung gian.

c. tng

Gi s M c cho dng ng (ngha l cc phn t ch gm ton t v ton hng).

Khi to stack: S.Duyt t M t tri qua phi:

- Nu gp ton hng th a vo ngn xp S.

- Nu gp ton t th ly hai phn t trn nh ngn xp, thc hin php ton ri a

kt qu tr li ngn xp.

- Thc hin cho n khi vt cn ht tt c phn t ca M.

d. Gii thut

CODE:

Input: M: Biu thc hu t.

1. Khi to stack S={}; // Stack S rng.

2. for (i=1;i


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e. V d

CODE:

Cu I, 2, thi tuyn sinh nm 2011.Input: M = 6 1 - 7 1 - * 11 1 + 3 / 2 + /Khi to S={};- M= 6, S={6};- M= 1, S={1};- M= -, S={5};- M= 7, S={5,7};- M= 1, S={5,7,1};- M= -, S={5,6};- M= *, S={30};- M= 11, S={30,11};- M= 1, S={30,11,1};- M= +, S={30,12};

- M= 3, S={30,12,3};- M= /, S={30,4};- M= 2, S={30,4,2};- M= +, S={30,6};- M= /, S={5};- M= , Dng, S={5};Output: Result=Top(S) = 5

[THUT TON - JAVA] CHUYN BIU THC TRUNG TSANG HU T JAVA CONVERTS INFIXTO POSTFIX

Cc biu thc i s c s dng hng ngy u c biu din di dng trung

t (infix). Cch biu din ny rt d hiu vi con ngi v hu ht cc ton t (+,

-, *, /) u l ton t hai ngi v chng phn cch gia hai ton hng vi nhau.

Tuy nhin i vi my tnh, tnh c gi tr ca mt biu thc i s theo

dng ny khng n gin nh ta vn lm. khc phc iu , my tnh cn

chuyn cch biu din cc biu thc i s t trung t sang mt dng khc l

tin t hoc hu t.

Th no l biu thc tin t, trung t v hu t

Trong on gii thiu trn c l bn cng hnh dung c th no l biu thc

trung t, hiu n gin tc l ton t s c t gia hai ton hng, d nhin


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y phi l ton t hai ngi. Vy da vo v tr ca ca ton t, liu ta c th biu

din biu thc i s di dng khc? Cu tr li l c, v nh ni, ta c ba

cch l biu thc tin t (prefix), trung t (infix) v hu t (postfix). Hy xem mt

cht gii thiu v cch biu din biu thc tin t v hu t hiu r hn v

chng.

- Prefix: Biu thc tin t c biu din bng cch t ton t ln trc cc ton

hng. Cch biu din ny cn c bit n vi tn gi k php Ba Lan do nh

ton hc Ba Lan Jan ukasiewicz pht minh nm 1920. Vi cch biu din ny,

thay v vit x+y nh dng trung t, ta s vit +xy. Ty theo u tin ca ton

t m chng s c sp xp khc nhau, bn c th xem mt s v d pha sau

phn gii thiu ny.

- Postfix: Ngc li vi cch Prefix, tc l cc ton t s c t sau cc ton

hng. Cch biu din ny c gi l k php nghch o Ba Lan hoc c

vit tt l RPN (Reverse Polish notation), c pht minh vo khong gia thp

k 1950 bi mt trit hc gia v nh khoa hc my tnh Charles Hamblin ngi

c.

Mt s v d:


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Phng php chuyn t biu thc trung t sang hu t

u tin ca cc ton t

Mt trong nhng iu quan trng trc khi bt u l phi tnh ton c u

tin ca cc ton t trong biu thc nhp vo. n gin ta ch xt cc ton t

hai ngi v thng dng bao gm: multiply (+),subtract (-), multiply (*), divide

(/). Theo cc ton t *, / c cng u tin v cao hn hai ton t +, -.

Nh vy ta c phng thc ly u tin ton t nh sau:

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publicintpriority(charc){ // thiet lap thu tu uu tienif(c == '+'|| c == '-') return1;elseif(c == '*'|| c == '/') return2;

elsereturn0;}

Kim tra ton t v ton hng

Trong thut ton chuyn i ny ta cn c cc phng thc kim tra xem mt

thnh phn ca chui c phi l ton t hoc ton hng khng. Thay v s dng

cc cu trc if hoc switch di dng v bt tin khi pht trin, ta s dng Regex

kim tra.

Ngoi ra v chui nhp vo l mt biu thc i s, nn cc ton hng ta s xt

khng ch l cc ch s m cn c ch ci t a-z v A-Z.

C mt quy tc na l khi dng ch ci th ch cho php duy nht mt ch ci i

din cho mt ton hng, cn khi dng ch s th c th nhiu ch s ghp thnh

mt ton hng.

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publicboolean

isOperator(char

c){ // kiem tra xem co phai toan tucharoperator[] = { '+', '-', '*', '/', ')', '('};

Arrays.sort(operator);if(Arrays.binarySearch(operator, c) > -1)

returntrue;elsereturnfalse;

}

Chun ha biu thc Infix trc khi chuyn i


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Cc biu thc Infix khi nhp vo c th d tha cc khong trng, cc k t khng

ph hp hoc vit sai c php.

Phn ny cc bn xem bi chun ha xu trong Java

Ngoi ra cc bn cn phi ghp cc ch s lin nhau thnh s (ton hng), tch

cc ton t, phn cch vi nhau bng mt khong trng. Cc phn t ny ti s

gi l mt token.

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publicString[] processString(String sMath){ // xu ly bieu thuc nhap vao thaString s1 = "", elementMath[] = null;InfixToPostfix IFP = newInfixToPostfix();sMath = sMath.trim();

sMath = sMath.replaceAll("\\s+"," "); // chuan hoa sMathfor(inti=0; i


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hoc bng ton t hin ti th ly ton t ra khi stack v cho ra

output.

+/a ton t hin ti vo stack

Sau khi duyt ht biu thc infix, nu trong stack cn phn t th ly

cc token trong ra v cho ln lt vo output.


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VD: Chng ta s chuyn biu thc A*B+C*((D-E)+F)/G t dng Infix sang dng

Postfix:


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Ci t trn Java

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publicString[] postfix(String[] elementMath){InfixToPostfix IFP = newInfixToPostfix();String s1 = "", E[];Stack S = newStack ();for(inti=0; i


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charc = elementMath[i].charAt(0); // c la ky tu dau tien cua m

if(!IFP.isOperator(c)) // neu c khong la toan tus1 = s1 + " "+ elementMath[i]; // xuat elem vao s1

else{ // c la toan tuif(c == '(') S.push(elementMath[i]); // c la "(" -> day pelse{

if(c == ')'){ // c la ")"charc1; //duyet lai cac phan tu trong Stackdo{

c1 = S.peek().charAt(0); // c1 la ky tu dau tif(c1 != '(') s1 = s1 + " "+ S.peek(); // trS.pop();

}while(c1 != '(');}else{

while(!S.isEmpty() && IFP.priority(S.peek().charAt(// Stack khong rong va trong khi phan tu trong Stack

s1 = s1 + " "+ S.peek(); // xuat phan tu tronS.pop();

}S.push(elementMath[i]); // dua phan tu hien tai vao

}}

}}while(!S.isEmpty()){ // Neu Stack con phan tu thi day het vao s1

s1 = s1 + " "+ S.peek();S.pop();

}E = s1.split(" "); // tach s1 thanh cac phan tureturnE;

}

V cui cng l chy chung trnh chnh:

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publicstaticvoidmain(String[] agrs){String sMath, elementMath[] = null;

InfixToPostfix IFP = newInfixToPostfix();Scanner input = newScanner(System.in);sMath = input.nextLine(); // nhap bieu thucelementMath = IFP.processString(sMath); // tach bieu thuc thanhelementMath = IFP.postfix(elementMath); // dua cac phan tu ve dfor(inti=0; i


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