栈的应用：表达式求值运算

Posted 2020-06-25 yoyo_sincerely

 

http://blog.csdn.net/wenzhou1219/article/details/23961307

#include <iostream>
#include <cstdio>
#include <stack>
#include <cstring>
#include <stdlib.h>
using namespace std;

#define LEVEL_invalid -1
#define LEVEL_same 0
#define LEVEL_smaller 1
#define LEVEL_bigger 2
stack <char> ops,mid_ops;
stack <double> vals,mid;
double m,n;
char operator_level_1[10] = {‘#‘, ‘(‘, ‘+‘, ‘-‘, ‘*‘, ‘/‘, ‘)‘};
char operator_level_2[10] = {‘#‘, ‘)‘, ‘-‘, ‘+‘, ‘/‘, ‘*‘, ‘(‘};

int compare_operator(char op1, char op2){
    int level1_index1 = -1,level1_index2 = -1;
    int level2_index1 = -1,level2_index2 = -1;
    if((op1 == ‘)‘ && op2 == ‘(‘) || (op1 == ‘#‘ && op2 == ‘)‘) || (op1 == ‘(‘ && op2 == ‘#‘))///)和(， #和)，(和#都是不能配对的，不合法；
        return LEVEL_invalid;
    if((op1 == ‘(‘ && op2 == ‘)‘) || (op1 == ‘#‘ && op2 == ‘#‘))
        return LEVEL_same;

    while(operator_level_1[++level1_index1] != op1); ///op1 在 level_1中的位置大小；
    while(operator_level_1[++level1_index2] != op2); ///op2 在 level_1中的位置大小；
    while(operator_level_2[++level2_index1] != op1); ///op1 在 level_2中的位置大小；
    while(operator_level_2[++level2_index2] != op2); ///op2 在 level_2中的位置大小；

    ///1.判断两个操作符的优先级关系
    ///2.‘#‘ < ‘+‘, ‘-‘ < ‘*‘, ‘/‘
    ///3.两个相同的运算符或同级别的运算符(‘+‘和‘-‘，‘*‘和‘/‘)
    ///对比时应判定为前一个运算符优先级别高,这样才能保证连续
    ///两个相同的运算符或同级别的运算符出现时前一个运算符出栈
    ///完成一次计算
    if((level1_index1 - level1_index2 <0 && level2_index1 - level2_index2 <0)||op1 == ‘(‘ || op2 == ‘(‘)
        return LEVEL_smaller;
    else
        return LEVEL_bigger;
}

void cal_operator(char op2){
    char op1 = ops.top();
    int result = compare_operator(op1, op2);
    if(result == LEVEL_smaller)
        ops.push(op2);
    else if(result == LEVEL_same)
        ops.pop();
    else if(result == LEVEL_bigger){
        char s = ops.top();ops.pop();
        double n,m;
        m = vals.top();vals.pop();
        n = vals.top();vals.pop();
        if(s == ‘*‘) vals.push(n * m);
        else if(s == ‘+‘) vals.push(n + m);
        else if(s == ‘/‘) vals.push(n / m);
        else if(s == ‘-‘) vals.push(n - m);
        if((op2 == ‘)‘ && ops.top() == ‘(‘) || (op2 == ‘#‘ && ops.top() == ‘#‘))
            ops.pop();
        else
            ops.push(op2);
    }
}

int main(){
    //freopen("in.txt","r",stdin);
    char x;
    ops.push(‘#‘);
    while((x = getchar()) != EOF){
        //cout<<x;
        if(((x > ‘9‘)||(x < ‘0‘)) && !mid.empty()) {
            vals.push(mid.top());
            while(!mid.empty())
                mid.pop();
        }
        else if(x >= ‘0‘ && x <= ‘9‘) {
            if(mid.empty()) {n = x - ‘0‘;mid.push(n);}
            else if(!mid.empty()){
                m = mid.top();
                n = m*10 + x - ‘0‘;
                mid.pop();
                mid.push(n);
            }
        }
        if(x == ‘\n‘) {
            ops.push(‘#‘);
            while (!ops.empty()){
                char s = ops.top();ops.pop();
                cal_operator(s);
            }
            break;
        }
        if(x > ‘9‘ || x < ‘0‘)
            cal_operator(x);
    }
    return 0;
}