动手敲代码——栈（经典问题练习）

/*用栈实现计算器*/
/*经典的用来练习栈的题目，需要借助逆波兰表达式*/

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

#define MAXSIZE 64
#define OK 1
#define ERROR 0

typedef struct
{
int data[MAXSIZE];
int top;
}stacknode;

/*创建一个顺序栈*/
stacknode * sqstack_create(void)
{
stacknode * p;

p = (stacknode *)malloc(sizeof(*p));
assert(p);      //分配空间后加上断言是个好习惯
p->top = -1;

return p;
}

/*入栈操作*/
int sqstack_push(int n, stacknode * p)
{
if(p->top == (MAXSIZE - 1)) //栈满
{
return ERROR;
}
p->data[++p->top] = n;

return OK;
}

/*出栈操作*/
int sqstack_pop(int *dest, stacknode * p)
{
if(p->top == -1)    //栈空
{
return ERROR;
}
*dest = p->data[p->top--];

return OK;
}

/*获取栈顶值*/
int sqstack_top(int *dest, stacknode *p)        //const修饰指针*p，p指向的内容不可改变
{
if(p->top == -1)    //栈空
{
return ERROR;
}
*dest = p->data[p->top];

return OK;
}

/*打印栈中内容*/
int sqstack_print(stacknode *p)
{
int n = p->top;

if(n == -1) //栈空
{
return ERROR;
}
while(n)
{
printf(" %d ",p->data[n--]);
}
printf(" %d ",p->data
);
printf("\n");
}

/*判断栈是否为空*/
int sqstack_is_empty(stacknode *p)
{
return (p->top == -1);
}

/*优先级判断函数*/
int get_priority(int ope)
{
switch (ope)
{
case '+':
case '-':return 1;
case '*':
case '/':return 2;
case '(':return 0;
default:return ERROR;
}
}

/*提取两个数和一个操作符进行计算，得到的结果再压栈*/
void compute(stacknode *pnum,int ope)
{
int num1,num2;
int result;

sqstack_pop(&num1,pnum);
sqstack_pop(&num2,pnum);
switch (ope)
{
case '+':   result = num1 + num2;
case '-':   result = num1 - num2;
case '*':   result = num1 * num2;
case '/':   result = num1 / num2;
}
sqstack_push(result,pnum);
}

/*进行优先级计算*/
void deal_ope(stacknode *pnum,stacknode *pope,int ope)
{
int old_ope;

if(sqstack_is_empty(pope)||ope == '(')  //当空栈或符号为（ 时，将符号压栈后返回
{
sqstack_push(ope,pope);
return;
}
sqstack_top(&old_ope,pope);     //得到栈顶元素
if(get_priority(ope)>get_priority(old_ope))             //比较优先级，入参的符号优先级大于栈顶元素时， 将符号压栈后返回
{
sqstack_push(ope,pope);
return;
}
while(get_priority(ope)<=get_priority(old_ope))
{
sqstack_pop(&old_ope,pope);
compute(pnum,old_ope);      //以栈顶符号计算 后压栈
if(sqstack_is_empty(pope))
{
break;
}
sqstack_top(&old_ope,pope);
}
sqstack_push(ope,pope); //将新的操作符入栈
}

//处理括号 ，进行括号里的高优先级计算，之后将计算结果压入数字栈，并且将括号丢弃
void deal_bracket(stacknode *pnum,stacknode *pope)
{
int old_ope;

sqstack_top(&old_ope,pope);     //得到符号栈栈顶元素
while(old_ope != '(')
{
sqstack_pop(&old_ope,pope);     //符号栈出栈
compute(pnum,old_ope);          //计算后将结果压入数字栈
sqstack_top(&old_ope,pope);     //得到栈顶元素，进行下一次循环
}
sqstack_pop(&old_ope,pope);     //将（ 出栈丢去
}

int main()
{
/*str为表达式数组*/
char str[MAXSIZE];
printf("请输入你要计算的表达式:\n");
gets(str);

int i = 0;
int value = 0;   //数字的值
int flag = 0;
int old_ope;

stacknode *pnum,*pope;      // 定义两个指向栈结构体的指针

pnum = sqstack_create();  // 创建存放数字的栈

pope = sqstack_create();  // 创建存放运算符号的栈

/* 表达式字符串解析函数,然后将高优先级的符号/(*)进行计算重新入栈
退出while大家的优先级都一样+-*/
while (str[i] != '\0')
{
//获取输入的数字
if (str[i] >= '0' && str[i] <= '9')//num
{
value = value * 10 + str[i] - '0';
flag = 1;
}
else//ope
{
if (flag)
{
//flag = 1说明value里面存储了数字，将其入栈
sqstack_push (value, pnum);
//num标志清零，value存放数字的变量清零
flag = 0;
value = 0;

}
if(str[i] == ')')
{
//如果是右括号，则
deal_bracket(pnum,pope);
}
else//+-*/(等情况
{
deal_ope(pnum,pope,str[i]);
}
}
i++;
}

//如果flag = 1.说明value里面还有数值,将其入栈

if (flag)
{
sqstack_push(value,pnum);
}
//然后将符号与数字依次出栈计算。数字出栈计算完成之后回再次在compute中入栈
while (!sqstack_is_empty(pope))
{
sqstack_pop(&old_ope,pope);
compute(pnum,old_ope);
}
//取出数字栈最后剩下的数字，就是最后的答案
sqstack_pop(&value,pnum);
//打印结果
printf("%s = %d\n",str,value);

return 0;
}