algo_156(蓝桥杯) 表达式计算

问题描述

　　输入一个只包含加减乖除和括号的合法表达式，求表达式的值。其中除表示整除。

输入格式

　　输入一行，包含一个表达式。

输出格式

　　输出这个表达式的值。

样例输入

1-2+3*(4-5)

样例输出

-4

数据规模和约定

　　表达式长度不超过100，表达式运算合法且运算过程都在int内进行。

#include<iostream>
#include<cstdio>
#include<stack>
#include<vector>
#include<cstring>
#include<cctype>
using namespace std;
//#define LOCAL 

struct Node {
    int key; // 为0时表示当前节点为操作数，为1表示为操作符；
    int number;
    char oper;
}a[105];

//存放拆解后的表达式
stack<Node> analytic;
stack<Node> calc;
//存放后序表达式
vector<Node> post;             

int main() {
#ifdef LOCAL
    freopen("algo_156.txt", "r", stdin);
#endif
    string str;
    int num_a = 0;
    cin >> str;
    //解决负数开头的问题
    if(str[0] == ‘-‘) {
        string str1 = "0";
        str = str1 + str;
    }
    //解析表达式
    int len = str.length();
    for(int i = 0; i < len; i++) {
        int num = 0;
        int temp = 0;
        while(str[i] != ‘\0‘ && isdigit(str[i])) {
            temp = 1;
            num *= 10;
            num += (str[i] - ‘0‘);
            i++;
        }
        if(temp == 1) {
            a[num_a].key = 0;
        a[num_a].number = num;
        post.push_back(a[num_a]);
        num_a++;
        }

        if(str[i] == ‘(‘) {
            a[num_a].key = 1;
            a[num_a].oper = ‘(‘;
            analytic.push(a[num_a]);
            num_a++;
        }
        else if(str[i] == ‘)‘) {
            while(!analytic.empty() && analytic.top().oper != ‘(‘) {
                post.push_back(analytic.top());
                 analytic.pop();
            }
            if(analytic.top().oper == ‘(‘)
                analytic.pop();
        }
        else if(str[i] == ‘+‘ || str[i] == ‘-‘) {
            while(!analytic.empty() && analytic.top().oper != ‘(‘) {
                post.push_back(analytic.top());
                 analytic.pop();
            }
            a[num_a].key = 1;
            a[num_a].oper = str[i];
            analytic.push(a[num_a]);
            num_a++;
        }
        else if(str[i] == ‘*‘ || str[i] == ‘/‘) {
            while(!analytic.empty() && (analytic.top().oper == ‘*‘ || analytic.top().oper == ‘/‘)) {
                post.push_back(analytic.top());
                 analytic.pop();
            }
            a[num_a].key = 1;
            a[num_a].oper = str[i];
            analytic.push(a[num_a]);
            num_a++;
        }
    }
    //将栈中元素释放
    while(!analytic.empty()) {
        post.push_back(analytic.top());
        analytic.pop();
    }
    //输出后序表达式
    /*
    for(int m = 0; m < post.size(); m++) {
        if(post[m].key == 1) cout << post[m].oper;
        else cout << post[m].number;
    }
    cout << endl;
    */
    //计算后序表达式
    for(int j = 0; j < post.size(); j++) {
        if(post[j].key == 0) {
            calc.push(post[j]);
        }
        else if(post[j].key == 1) {
            int s, left, right;
            right = calc.top().number;
            calc.pop();
            left = calc.top().number;
            switch(post[j].oper) {
                case ‘+‘ : s = left + right;break;
                case ‘-‘ : s = left - right;break;
                case ‘*‘ : s = left * right;break;
                case ‘/‘ : s = left / right;break;
                default : return -1;
            }
            calc.top().number = s;
        }
    }
    int sum = calc.top().number;
    cout << sum << endl;
    return 0;
}

原文地址：https://www.cnblogs.com/bearcarl/p/8527573.html