《数据结构》严蔚敏 算法3.4 用栈实现表达式求值

emmmm 我又偷懒了，看了一下书上的算法，总感觉不是很好的算法，我觉得可以类比前面的符号匹配来进行表达式求值，但是今天有点不想写了，先copy一个答案吧

原文链接：https://blog.csdn.net/hello_sheep/article/details/75635777

输入
以“#”结尾的表达式，运算数为正整数。每个表达式占一行。

输出
输出表达式运算的结果。

置运算符栈为空栈，表达式的起始符’#‘为栈底元素
依次读入表达式的每个字符，
若是操作数则进OPND栈，
若是运算符则和OPTR栈的栈顶元素比较优先权后进行相应操作，
直至整个表达式求值完毕(即OPTR栈的栈顶元素和当前读入的字符均为’#’）

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
 
 
#define N 1000+10
#define STACK_INIT_SIZE 100
#define STACKINCREMENT 10
#define OK 1
#define OVERFLOW 0
#define ERROR 0
 
char str[N];
typedef  int Status;
typedef  int SElemType;
 
typedef struct{
	SElemType *base;
	SElemType *top;
	int stacksize;
}SqStack;
 
unsigned char prior[7][7] = {
{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=',' '},
{'<','<','<','<','<',' ','>'},
{'<','<','<','<','<',' ','='}};
 
char OPSET[7] = {'+','-','*','/','(',')','#'};
 
Status InitStack(SqStack *s)//初始化栈 
{
	s->base = (SElemType*)malloc(STACK_INIT_SIZE*sizeof(SElemType));
	if(!s->base)
		exit(OVERFLOW);
	s->top = s->base ;
	s->stacksize = STACK_INIT_SIZE;
	return OK;
}
 
Status Push(SqStack *s,SElemType c)//入栈 
{
	if((s->top - s->base ) >= s->stacksize )
	{
		s->base = (SElemType*)realloc(s->base ,(s->stacksize +STACKINCREMENT)*sizeof(SElemType));
		if(!s->base )
			exit(OVERFLOW);
		s->stacksize += STACKINCREMENT;
	}
	*(s->top)++ = c;
	return OK;
}
 
Status GetTop(SqStack *s)//取栈顶元素 
{
	SElemType e;
	if(s->base == s->top )
		return ERROR;
	e = *(s->top-1)    ;
	return e;
}
 
Status In(char c,char str[])//判断是否为运算符 
{
	int i = 0;
	while(c != str[i])
	{
		i++;
	}
	if(i < 7)
		return OK;
	return ERROR;
}
 
void  Strcat(char *str1,char *str2)//字符串连接函数，把字符串str2连接到str1后 
{
	int i = 0, j = 0;
	while(str1[i]!='\0')
	{
		i++;
	}
	while(str2[j]!='\0')
	{
		str1[i++] = str2[j++];
	}
	str1[i] = '\0';
} 
 
Status Atoi(char *c)//把字符串转为数字 
{
	int data= 0,d = 0;
	int i = 0;
	while(c[i]!='\0')
	{
		data = data*10 + c[i]-'0';
		i++;
	}
	return data;	
} 
 
Status precede(int a,char b)//判断优先级函数 
{
	int i = 0,j = 0;
	while(OPSET[i] != a)
	{
		i++;
	}
	while(OPSET[j] != b)
	{
		j++;
	}
	return prior[i][j];
}
 
Status Pop(SqStack *s)//脱括号函数 
{
	int e;
	if(s->base == s->top )
		return ERROR;
	e = *--(s->top);
	return e;
}
 
Status Opereta(int a,int b,int c)//运算函数 
{
	switch(b)
	{
		case '+':
			return a+c;
		case '-':
			return a-c;
		case '*':
			return a*c;
		case '/':
			return a/c;
	} 

	return OK;
}
 
int EvaluateExpression(char *MyExpression)//算法3.4 
{//算术表达式求值的算符优先算法。
//设OPTR和OPND分别为运算符栈和运算数栈
	SqStack OPTR;//运算符栈，字符元素 
	SqStack OPND;//运算数栈，实数元素 
	
	char TempData[20];
	int data,a,b;
	char *c,Dr[2],e;
	int theta;
	
	InitStack(&OPTR);
	Push(&OPTR,'#');
	InitStack(&OPND);
	
	c = MyExpression;
	TempData[0] = '\0';
	while(*c != '#'|| GetTop(&OPTR) != '#')
	{
		
		if(!In(*c,OPSET))//不是运算符则进栈 
		{
			Dr[0] = *c;
			Dr[1] = '\0';
			Strcat(TempData,Dr);
			c++; 
			if(In(*c,OPSET))//是运算符时 
			{
				data = Atoi(TempData);
				Push(&OPND,data);
				TempData[0] = '\0';
			}
		}
		else
		{
			switch(precede(GetTop(&OPTR),*c))
			{
				case '<':
					Push(&OPTR,*c);
					c++;
					break;
				case '=':
					Pop(&OPTR);
					
					c++;
					break;
				case '>':
					a = Pop(&OPND);
					b = Pop(&OPND);
					theta = Pop(&OPTR);
					Push(&OPND,Opereta(b,theta,a));
					break;
			}
		}
	}
	
	return GetTop(&OPND);
}
int main()
{
	
	while(scanf("%s",str)!=EOF)
	{
		printf("%d\n",EvaluateExpression(str));
	}
	return 0;
 }