数据结构专项训练-栈

求前缀表达式的值 

算术表达式有前缀表示法、中缀表示法和后缀表示法等形式。前缀表达式指二元运算符位于两个运算数之前，例如2+3*(7-4)+8/4的前缀表达式是：+ + 2 * 3 - 7 4 / 8 4。请设计程序计算前缀表达式的结果值。

输入格式:

输入在一行内给出不超过30个字符的前缀表达式，只包含+、-、*、/以及运算数，不同对象（运算数、运算符号）之间以空格分隔。

输出格式:

输出前缀表达式的运算结果，保留小数点后1位，或错误信息"ERROR"。

样例输入与输出：

序号	输入	输出
1	+ + 2 * 3 - 7 4 / 8 4	13.0
2	/ -25 + * - 2 3 4 / 8 4	12.5
3	/ 5 + * - 2 3 4 / 8 2	ERROR
4	+10.23	10.2

这里就不按照惯例给出建议的测试用例了，直接用题目中给的4个就可以了。

下面给出代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXSIZE 30

typedef double DateType;

typedef struct
{
    DateType date[MAXSIZE];
    int top;
}Zhan, *PZ;

void Push(PZ S, DateType n)
{
    S->top++;
    S->date[S->top] = n;
}

DateType Pop(PZ S)
{
    DateType m = S->date[S->top];
    S->top--;
    return m;
}

int main()
{
    Zhan Z;
    PZ p = &Z;
    p->top = -1;
    char str[31];
    double num;
    gets(str);
    int i;
    int size = strlen(str);
    for (i = size - 1; i >= 0; i--)
    {
        if (str[i] >= '0'&&str[i] <= '9')
        {
            num = str[i] - '0';
            int k = 10;
            for (i = i - 1; i >= 0; i--)
            {
                if ((str[i] >= '0'&&str[i] <= '9') || str[i] == '.')
                {
                    if (str[i] >= '0'&&str[i] <= '9')
                    {
                        num = num + (str[i] - '0')*k;
                        k = k * 10;
                    }
                    else
                    {
                        num /= k;
                        k = 1;
                    }
                }
                else if (str[i] == '-')
                {
                    num = -num;
                }
                else break;
            }
            Push(p, num);
        }
        else if (str[i] == ' ')
        {
            continue;
        }
        else
        {
            if (p->top >= 1)
            {
                double a, b, s;
                a = Pop(p);
                b = Pop(p);
                if (str[i] == '+')
                {
                    s = a + b;
                }
                else if (str[i] == '-')
                {
                    s = a - b;
                }
                else if (str[i] == '*')
                {
                    s = a * b;
                }
                else
                {
                    if (b == 0)
                    {
                        printf("ERROR
");
                        return 0;
                    }
                    else s = a / b;
                }
                Push(p, s);
            }
            else
            {
                printf("ERROR
");
                return 0;
            }
        }
    }
    if (p->top == 0) printf("%.1f
", p->date[p->top]);
    else printf("ERROR
");
    system("pause");
    return 0;
}

写的比较好的就是在一个for循环中再嵌套一个for循环，而且变量还是i

Pop Sequence 

Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.

Sample Input:

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:

YES
NO
NO
YES
NO

说明：该题目就是要求1到N依次入栈，出栈随机，而得到与输入相同的队列，如果能得到，则输出YES，否则输出NO

代码如下：

#include <stdio.h>
#include <stdlib.h>
#define MAXSIZE 1000

typedef int DateType;
typedef struct
{
    DateType date[MAXSIZE];
    int top;
}Zhan,*PZ; 

void Push(PZ S,char n)//入栈
{
    S->top++;
    S->date[S->top] = n;
}

void Pop(PZ S)//出栈
{
    S->top--;
}


int main()
{
    int m,n,k,i;
    scanf("%d %d %d",&m, &n, &k); //m为栈的最大值，n为输出的队列长度，k为测试样例个数
    for (i = 0; i < k; i++)
    {
        Zhan Z; 
        PZ p = &Z;
        p->top = -1; //对栈定义及初始化
        int a[1000]; //用来存放每组样例
        int g;
        for (g = 0; g < n; g++)   //初始化每个样例
        {
            scanf("%d", &a[g]);
        }
        int t = 0, num = 1;  //t用来检索数组a当前的数据位置，num为依次入栈的自然数据
        while (t < n)                   
        {
            if (a[t] == num && p->top < m - 1) //如果当前自然数据等于当前检索的样例的a[t]值，且栈未满，即可通过入栈再出栈完成当前自然数字
            {
                t++;
                num++;
            }
            else if (p->top > -1 && p->top < m - 1 && a[t] == p->date[p->top]) //如果栈顶数字等于检索样例a[t]的值，则可出栈
            {
                Pop(p);
                t++;
            }
            else if (num != a[t] && p->top < m - 1) //如果当前自然数据不等于当前检索的a[t]，且栈未满,则将该自然数据入栈
            {
                Push(p, num);
                num++;
            }
            else break; //若上述的情况均不成立，则说明该样例已经不是正确的，可以结束检索，退出循环
        }
        if (t < n) printf("NO
");//检索没进行完，循环中途退出，说明样例不正确
        else printf("YES
");
    }
    system("pause");
    return 0;
}