在一個數組中實現兩個堆疊
阿新 • • 發佈:2018-12-20
函式介面定義:
Stack CreateStack( int MaxSize );
bool Push( Stack S, ElementType X, int Tag );
ElementType Pop( Stack S, int Tag );
其中Tag是堆疊編號,取1或2;MaxSize堆疊陣列的規模;Stack結構定義如下:
typedef int Position; struct SNode { ElementType *Data; Position Top1, Top2; int MaxSize; }; typedef struct SNode *Stack;
注意:如果堆疊已滿,Push函式必須輸出“Stack Full”並且返回false;如果某堆疊是空的,則Pop函式必須輸出“Stack Tag Empty”(其中Tag是該堆疊的編號),並且返回ERROR。
裁判程式樣例:
#include <stdio.h> #include <stdlib.h> #define ERROR 1e8 typedef int ElementType; typedef enum { push, pop, end } Operation; typedef enum { false, true } bool; typedef int Position; struct SNode { ElementType *Data; Position Top1, Top2; int MaxSize; }; typedef struct SNode *Stack; Stack CreateStack( int MaxSize ); bool Push( Stack S, ElementType X, int Tag ); ElementType Pop( Stack S, int Tag ); Operation GetOp(); /* details omitted */ void PrintStack( Stack S, int Tag ); /* details omitted */ int main() { int N, Tag, X; Stack S; int done = 0; scanf("%d", &N); S = CreateStack(N); while ( !done ) { switch( GetOp() ) { case push: scanf("%d %d", &Tag, &X); if (!Push(S, X, Tag)) printf("Stack %d is Full!\n", Tag); break; case pop: scanf("%d", &Tag); X = Pop(S, Tag); if ( X==ERROR ) printf("Stack %d is Empty!\n", Tag); break; case end: PrintStack(S, 1); PrintStack(S, 2); done = 1; break; } } return 0; } /* 你的程式碼將被嵌在這裡 */
輸入樣例:
5
Push 1 1
Pop 2
Push 2 11
Push 1 2
Push 2 12
Pop 1
Push 2 13
Push 2 14
Push 1 3
Pop 2
End
輸出樣例:
Stack 2 Empty
Stack 2 is Empty!
Stack Full
Stack 1 is Full!
Pop from Stack 1: 1
Pop from Stack 2: 13 12 11
正確程式碼:(PTA通過)
/*THE ANSWER IS:*/
Stack CreateStack( int MaxSize )
{
Stack S;
S=(Stack)malloc(sizeof(struct SNode));
S->MaxSize=MaxSize;
S->Top1=-1;
S->Top2=S->MaxSize;
S->Data=(ElementType *)malloc(S->MaxSize*sizeof(ElementType));
return S;
}
bool Push( Stack S, ElementType X, int Tag )
{
if(Tag==1)
{
if(S->Top1+1==S->Top2)/*如果棧滿*/
{
printf("Stack Full");
return false;
}else{
S->Data [++(S->Top1)]=X;
return true;
}
}else if(Tag==2)
{
if(S->Top1+1==S->Top2)/*如果棧滿*/
{
printf("Stack Full");
return false;
}else{
S->Data [--(S->Top2)]=X;
return true;
}
}
}
ElementType Pop( Stack S, int Tag )
{
if(Tag==1)
{
if(S->Top1==-1)/*棧空*/
{
printf("Stack %d Empty",Tag);
return ERROR;
}else{
return S->Data[(S->Top1)--];
}
}else if(Tag==2)
{
if(S->Top2==S->MaxSize)
{
printf("Stack %d Empty",Tag);
return ERROR;
}else{
return S->Data[(S->Top2)++];
}
}
}
SOME TIPS:
①:要注意題目要求,不要急於code。格式上的要求(printf(" \n")),看清輸出語句是讓你寫的還是程式自動執行的,
返回要求,明確要求返回false /ERROR. (ps:以後不要在這樣的問題山浪費時間!)
②:判斷棧空的條件:棧1空》》S->Tag1==-1;棧2空》》S->Tag2==MaxSize
判斷棧滿的條件:S->Tag1+1==S->Tag2
③:ElementType Pop( Stack S, int Tag )函式返回格式為ElementType,所以請眉毛下邊的兩位認真一些,謝謝!