数据结构复习笔记(2)

1，  若入栈的元素为n,则可得到的输出序列数量为 (2n)!/(n+1)(n!)(n!)。

2，  用两个长度相同的栈S1,S2构造一个队列。在S1中进行入队操作，S2中进行出队操作 ，判断队列空的条件是，S1和S2同时为空，判断队列满的条件是S1和S2同时为满。

void  EnQueue(ElemType x)
{
    if(!Full(S1))
    {//S1未满直接进入
        S1.Push(x);
    }
    else
    {
        if(Empty(S2)==true)
        {
            while(!Empty(S1))
            {
                S2.Push(S1.Pop());
            }
            S1.Push(x);        
        }
    }
}

ElemType DeQueue()
{
    if(!Empty(S2))
    {
        return S2.Pop();
    }
    else
    {
        if(!Empty(S1))
        {
            while(!Empty(S1))
            {
                S2.Push(S1.Pop());
            }
            return S2.Pop();
        }
        
    }
}

3.求两个正整数的最大公约数的非递归算法。

#define  MAX 100
struct  Stack
{
    int s;
    int t;
} ;

int  gcd( int  m, int  n)
{
    int k;
    Stack S[MAX];
    int top = -1,tmp;
    if(m<n)
    {
        tmp = m;
        m = n;
        n = tmp;
    }
    top++;
    S[top].s = m;
    S[top].t = n;
    while(top>=0&&S[top].t!=0)
    {
        if(S[top].t!=0)
        {
            tmp = S[top].s;
            m = S[top].t;
            n = m%tmp;
            top++;
            S[top].s = m;
            S[top].t = n;
        }
    }
    return S[top].s;
}

4．

                      n+1,m =0

Akm(m,n)  =           Akm(m-1,1) m!=0,n=0

                      Akm(m-1,Akm(m,n-1)),m!=0,n!=0

写非递归算法。

#define  MAXSIZE 100
typedef  struct
{
    int tm;
    int tn;
} Stack;

int  akm( int  m, int  n)
{
    Stack S[MAXSIZE];
    int top = 0;
    S[top].tm = m;
    S[top].tn = n;
    do
    {
        while(S[top].tm!=0)
        {
            while(S[top].tn!=0)
            {
                top++;
                S[top].tm = S[top-1].tm;
                S[top].tn = S[top-1].tn-1;
            }
            S[top].tm--;
            S[top].tn=1;
        }
        if(top>0)
        {
            top--;
            S[top].tm--;
            S[top].tn = S[top].tn+1;
        }
        
    }while(top!=0 || S[top].tm!=0);
    top--;
    return S[top+1].tn+1;
}

5．写出和下列递归过程等价的非递归过程

void  test( int   & sum)
{
    int k;
    scanf("%d",&k);
    if(k==0) sum = 1;
    else
    {
        test(sum);
        sum = k*sum;
    }
    printf("%d",sum);
}

分析：程序功能是按照输入的整数，按相反顺序进行累计乘法运算

#define  MAXSIZE 100
void  test( int   & sum)
{
    int Stack[MAXSIZE];
    int top = -1,k;
    sum = 1;
    scanf("%d",&k);
    while(k!=0)
    {
        Stack[++top] = k;
        scanf("%d",&k);
    }
    printf("%d",sum);
    while(top!=-1)
    {
        sum *=Stack[top--];
        printf("%d",sum);
    }
}
        

6．假设表达式由单字母变量和双目四则运算算符构成，写一个算法，将一个书写正确的表达式转换为逆波兰式。

void  ConPoland( char  express[], char  suffix[])
{
    char *p = express,ch1 = *p,ch2;
    int k = 0;
    InitStack(S);
    Push(S,'#');
    while(!StackEmpty(S))
    {
        if(!IsOperator(ch1))
            suffix[k++] = ch1;
        else
        {
            switch(ch1)
            {
                case '(':    
                    Push(S,ch1);break;
                case ')':    
                    Pop(S,ch2);
                    while(ch2!='(')    
                    {
                        suffix[k++] = ch2;
                        Pop(S,ch2);
                    }
                    break;
                default:
                    while(GetTop(S,ch2)&&precede(ch2,ch1))
                    {
                        suffix[k++] = ch2;
                        Pop(S,ch2);
                    }
                    if(ch1!='#')
                        Push(S,ch1);
                    break;                            
                    
            }
        }
        if(ch1!="#')    
            ch1 = *++p;
    }
    suffix[k] = '\0';
}