汉诺塔问题

递归实现

#include <iostream>
#include <stack>

using namespace std;

#define mvh(s, o)  cout << (s) << "=>" << (o) << endl;


void Hanoi(char a, char b, char c, int n)
{
    if (n < 1)
        return;
    Hanoi(a, c, b, n - 1);
    mvh(a, c)
    Hanoi(b, a, c, n - 1);
}

非递归实现

typedef struct HState
{
    char a;
    char b;
    char c;
    char flag;
    int  n;
} HState;

//非递归实现
void HanoiNo(char A, char B, char C, int n)
{
    stack<HState> HSS;

    HState hs1;
    hs1.a = A;
    hs1.b = B;
    hs1.c = C;
    hs1.flag = 0;
    hs1.n = n;

    HSS.push(hs1);

    while (!HSS.empty())
    {
        if (HSS.top().n < 1)
        {
            HSS.pop();
            if (!HSS.empty())
                HSS.top().flag ++;
        }
        else if (HSS.top().flag == 0)
        {
            HState hs;
            hs.a = HSS.top().a;
            hs.b = HSS.top().c;
            hs.c = HSS.top().b;
            hs.n = HSS.top().n - 1;
            hs.flag = 0;
            HSS.push(hs);
        }
        else if (HSS.top().flag == 1)
        {
            mvh(HSS.top().a, HSS.top().c)
            HState hs;
            hs.a = HSS.top().b;
            hs.b = HSS.top().a;
            hs.c = HSS.top().c;
            hs.n = HSS.top().n - 1;
            hs.flag = 0;
            HSS.push(hs);
        }
        else if (HSS.top().flag == 2)
        {
            HSS.pop();
            if (!HSS.empty())
                HSS.top().flag ++;
        }
        else
        {
            cout << "error..." << endl;
            break;
        }
    }
}

测试

#include <iostream>
#include <stack>

using namespace std;

int main()
{
    HanoiNo('A', 'B', 'C', 15);
    cout << "-------------------------------------"    << endl;
    Hanoi('A', 'B', 'C', 15);
    return 0;
}