算法导论 10.1-2 用一个数组实现两个栈

一、题目

    用一个数组A[ 1....N ]实现两个栈,除非数组的每一个单元都被使用,否则栈例程不能有溢出,注意PUSH和POP操作的时间应为O(1)。

二、解法

    对于一个数组,由它的两端作为栈底,栈向数组中间扩展。当数组中每个元素被用到时,栈满。

三、代码

struct Node;
typedef Node *ComStack;

struct Node
{
    int Capacity;
    int TopL;
    int TopR;
    ElementType *Array;
};


ComStack CreatComStack( int MaxElements );
int IsEmpty_L( ComStack S );
int IsEmpty_R( ComStack S );
int IsFull( ComStack S );
void MakeEmpty( ComStack S );
void Push_L( ElementType X, ComStack S );
void Push_R( ElementType X, ComStack S );
ElementType Pop_L( ComStack S );
ElementType Pop_R( ComStack S );
void DisposeComStack( ComStack S );


ComStack CreatComStack( int MaxElements )
{
    ComStack S;

    S = (ComStack)malloc( sizeof(struct Node) );
    if ( S == NULL )
    {
        printf( "Out of space" );
        return NULL;
    }

    S->Array = (ElementType *)malloc( sizeof(ElementType) * MaxElements );
    if ( S->Array == NULL )
    {
        printf( "Out of space" );
        return NULL;
    }

    S->Capacity = MaxElements;
    MakeEmpty(S);

    return S;
}


int IsEmpty_L( ComStack S )
{
    if ( S->TopL == -1 )
        return true;
    else
        return false;
}

int IsEmpty_R( ComStack S )
{
    if ( S->TopR == S->Capacity )
        return true;
    else
        return false;
}

int IsFull( ComStack S )
{
    if ( S->TopL + 1 == S->TopR )
        return true;
    else
        return false;
}

void MakeEmpty( ComStack S )
{
    S->TopL = -1;
    S->TopR = S->Capacity; // Capacity在数组界外
}

void Push_L(  ElementType X, ComStack S )
{
    if ( IsFull(S) )
        printf( "Stack is full" );
    else
    {
        S->TopL++;
        S->Array[S->TopL] = X;
    }
}

void Push_R( ElementType X, ComStack S )
{
    if ( IsFull(S) )
        printf( "Stack is full" );
    else
    {
        S->TopR--;
        S->Array[S->TopR] = X;
    }
}

ElementType Pop_L( ComStack S )
{
    ElementType TmpCell;

    if ( IsEmpty_L(S) )
        printf( "Left Stack is empty" );
    else
    {
        TmpCell = S->Array[S->TopL];
        S->TopL--;
    }

    return TmpCell;
}

ElementType Pop_R( ComStack S )
{
    ElementType TmpCell;

    if ( IsEmpty_R(S) )
        printf( "Right stack is empty" );
    else
    {
        TmpCell = S->Array[S->TopR];
        S->TopR++;
    }

    return TmpCell;
}

void DisposeComStack( ComStack S )
{
    if ( S != NULL )
    {
        free( S->Array );
        free( S );
    }
}
原文地址:https://www.cnblogs.com/tallisHe/p/4007252.html