數據結構：兩棧共享存儲空間

數組有兩個端點，兩個棧有兩個棧底，讓一個棧的棧底爲數組的始端，即下標爲0處，另一個棧爲棧的末端，即下標爲數組長度n-1處。這樣，如果兩個棧增加元素，就是兩端點向中間延伸。當top1 + 1 == top2 的時候爲棧滿。

示例代碼：（改編自《大話數據結構》）

#include <iostream>
using namespace std;

#define  MAXSIZE 20

typedef int ElemType;
typedef struct
{
    ElemType data[MAXSIZE];
    int top1; //棧1棧頂指針
    int top2; //棧2棧頂指針
} SqStack;

/* 構造一個空棧*/
bool InitStack(SqStack *Sq)
{
    cout << "Init Stack ..." << endl;
    Sq->top1 = -1; //表示空棧
    Sq->top2 = MAXSIZE;

    return true;
}
/* 置爲空棧 */
bool ClearStack(SqStack *Sq)
{
    cout << "Clear Stack ..." << endl;
    Sq->top1 = -1;
    Sq->top2 = MAXSIZE;
    return true;
}

bool StackEmpty(SqStack Sq)
{
    if (Sq.top1 == -1 && Sq.top2 == MAXSIZE)
        return true;
    else
        return false;
}

int StackLength(SqStack Sq)
{
    cout << "Stack Length : ";
    return Sq.top1 + 1 + MAXSIZE - Sq.top2;
}
/* 返回棧頂元素 */
bool GetTop(SqStack Sq, ElemType *ptr, int StackNum)
{
    if (StackNum == 1)
    {
        if (Sq.top1 != -1)
        {
            *ptr = Sq.data[Sq.top1];
            cout << "Get Top1 Item " << *ptr << endl;
            return true;
        }
        return false;
    }

    else if (StackNum == 2)
    {
        if (Sq.top2 != MAXSIZE)
        {
            *ptr = Sq.data[Sq.top2];
            cout << "Get Top2 Item " << *ptr << endl;
            return true;
        }
        return false;
    }
    else
    {
        cout << "Stack Num must be 1 or 2!" << endl;
        return false;
    }
}

/* 壓棧 */
bool Push(SqStack *Sq, ElemType Elem, int StackNum)
{
    if (StackNum == 1)
    {
        cout << "Push Item to Stack1 : " << Elem << endl;
        if (Sq->top1 + 1 == Sq->top2)   //棧滿
            return false;
        Sq->data[++Sq->top1] = Elem;
        return true;
    }
    else if (StackNum == 2)
    {
        cout << "Push Item to Stack2 : " << Elem << endl;
        if (Sq->top1 + 1 == Sq->top2)
            return false;
        Sq->data[--Sq->top2] = Elem;
        return true;
    }
    else
    {
        cout << "Stack Num must be 1 or 2!" << endl;
        return false;
    }
}
/* 出棧 */
bool Pop(SqStack *Sq, ElemType *ptr, int StackNum)
{
    if (StackNum == 1)
    {
        if (Sq->top1 == -1)
            return false;

        *ptr = Sq->data[Sq->top1--];
        cout << "Pop Item from Stack1 :  " << *ptr << endl;
        return true;
    }
    else if (StackNum == 2)
    {
        if (Sq->top2 == MAXSIZE)
            return false;
        *ptr = Sq->data[Sq->top2++];
        cout << "Pop Item from Stack2 :  " << *ptr << endl;
        return true;
    }
    else
    {
        cout << "Stack Num must be 1 or 2!" << endl;
        return false;
    }

}

bool StackTraverse(SqStack Sq)
{
    cout << "Traverse Stack ..." << endl;
    if (StackEmpty(Sq))
        return false;
    cout << "Stack1 : ";
    for (int i = 0; i <= Sq.top1; i++)
        cout << Sq.data[i] << ' ';
    cout << endl;
    cout << "Stack2 : ";
    for (int i = MAXSIZE - 1; i >= Sq.top2; i--)
        cout << Sq.data[i] << ' ';
    cout << endl;

    return true;
}

int main(void)
{
    SqStack Sq;
    InitStack(&Sq);
    for (int i = 0; i < 5; i++)
        Push(&Sq, i, 1);
    for (int i = 5; i < 10; i++)
        Push(&Sq, i, 2);
    StackTraverse(Sq);
    int result;
    Pop(&Sq, &result, 1);
    Pop(&Sq, &result, 2);
    StackTraverse(Sq);
    GetTop(Sq, &result, 1);
    GetTop(Sq, &result, 2);
    if (!StackEmpty(Sq))
        cout << StackLength(Sq) << endl;

    ClearStack(&Sq);

    return 0;
}

輸出爲：

事實上 ，使用這樣的數據結構，通常都是當兩個棧的空間需求有想法關係時，也就是當一個棧增長時另一個棧在縮短的情況。
還需要注意的一點是必須是同種數據類型的棧，否則不但不能更好地解決問題，反而會使問題更加複雜。

轉載自：http://blog.csdn.net/jnu_simba/article/details/8837121