逆波兰表达式求解

终于搞定逆波兰算法~！13 - 12 - 1 /*
* 缺点:
* 1:运算符比较少+ - * / ( )
* 2:只能运算10以下的数
* 3:运算结果为整数
*/
#include <stdio.h>//逆波兰
#define MAX 100
#define LEN 2 * MAX
//符号优先级表
int val[7][7] = {// + - * / ( ) #
/* 0 + */{ 1, 1, -1, -1, -1, 1, 1},
/* 1 - */{ 1, 1, -1, -1, -1, 1, 1},
/* 2 * */{ 1, 1, 1, 1, -1, 1, 1},
/* 3 / */{ 1, 1, 1, 1, -1, 1, 1},
/* 4 ( */{-1, -1, -1, -1, -1, 0, 8},
/* 5 ) */{ 1, 1, 1, 1, 8, 1, 1},
/* 6 # */{-1, -1, -1, -1, -1, 8, 0}
};
typedef char InputData;
typedef char ElemTypeOp;
typedef int ElemTypeS;
//-------------栈的操作-------------------
typedef struct Stack {//双栈
ElemTypeOp opData[MAX];//符号栈
ElemTypeS sData[MAX];//操作数栈
int topOp;//符号栈栈顶指针
int topS;//操作数栈顶指针
}Stack;
int initStack(Stack *s) {//初始化栈
s->opData[0] = '#';
s->topOp = 0;
s->topS = -1;
return 0;
}
int pushOp(Stack *s, ElemTypeOp op) {//运算符入栈
s->opData[s->topOp + 1] = op;
s->topOp++;
return 0;
}
int pushS(Stack *s, ElemTypeS num) {//操作数入栈
s->sData[s->topS + 1] = num;
s->topS++;
return 0;
}
int popOp(Stack *s, ElemTypeOp *op) {//运算符出栈
(*op) = s->opData[s->topOp];
s->topOp--;
return 0;
}
int popS(Stack *s, ElemTypeS *num) {//操作数出栈
(*num) = s->sData[s->topS];
s->topS--;
return 0;
}
ElemTypeOp getOpTop(Stack *s) {//得到运算符栈顶元素
return s->opData[s->topOp];
}
ElemTypeS getSTop(Stack *s) {//得到操作数栈顶元素
return s->sData[s->topS];
}
//---------------------------------
int valueOperator(ElemTypeOp op1, ElemTypeOp op2) {//取得符号优先级int x, y;
switch(op1) {
case '+': x = 0; break;
case '-': x = 1; break;
case '*': x = 2; break;
case '/': x = 3; break;
case '(': x = 4; break;
case ')': x = 5; break;
case '#': x = 6; break;
};
switch(op2) {
case '+': y = 0; break;
case '-': y = 1; break;
case '*': y = 2; break;
case '/': y = 3; break;
case '(': y = 4; break;
case ')': y = 5; break;
case '#': y = 6; break;
};
return val[x][y];
}
ElemTypeS calculate(ElemTypeOp op, ElemTypeS s1, ElemTypeS s2) {//两个操作数一个运算符计算值
ElemTypeS result;
switch(op) {
case '+': result = s1 + s2; break;
case '-': result = s1 - s2; break;
case '*': result = s1 * s2; break;
case '/': result = s1 / s2; break;
};
return result;//返回计算结果
}
int polish() {
int i = 0;
InputData input[LEN];//定义一个输入的表达式容器
ElemTypeOp tempOp;//临时的运算符
ElemTypeS tempS1, tempS2;//临时的操作数
Stack s;//定义一个运算要用的双栈
initStack(&s);//一定要初始化
printf("请输入数字小于10的多项式数式，在输入结尾加#字符:\n");
scanf("%s", input);
do {
if(input[i] >= '0' && input[i] <= '9')
{
pushS(&s, int(input[i] - '0'));
i++;
}
else if(valueOperator(getOpTop(&s), input[i]) == -1) {
pushOp(&s, input[i]);
i++;
}
else if(valueOperator(getOpTop(&s), input[i]) == 1) {
popOp(&s, &tempOp);
popS(&s, &tempS1);
popS(&s, &tempS2);
//printf("--------------%d\n", calculate(tempOp, tempS2, tempS1));
pushS(&s, calculate(tempOp, tempS2, tempS1));
}
else if(getOpTop(&s) == '(' && input[i] == ')') {
popOp(&s, &tempOp);
i++;
}
}while(!(getOpTop(&s) == input[i] && getOpTop(&s) == '#' && input[i] == '#')); return getSTop(&s);
}
int main() {
printf("answer: %d\n", polish());
return 0;
}。