四则运算/华为机试(C/C++)

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题目描述

请实现如下接口

    /* 功能：四则运算

     * 输入：strExpression：字符串格式的算术表达式，如: "3+2*{1+2*[-4/(8-6)+7]}"

         * 返回：算术表达式的计算结果

     */

    public static int calculate(String strExpression)

    {

        /* 请实现*/

        return 0;

    } 

约束：

1. pucExpression字符串中的有效字符包括[‘0’-‘9’],‘+’,‘-’, ‘*’,‘/’ ,‘(’， ‘)’,‘[’, ‘]’,‘{’ ,‘}’。

2. pucExpression算术表达式的有效性由调用者保证; 

输入描述:

输入一个算术表达式

输出描述:

得到计算结果

示例1

输入

3+2*{1+2*[-4/(8-6)+7]}

输出

25

代码：借用

#include<iostream>
#include<string>
#include<vector>
#include<stack>
using namespace std;
int main() 
{string s;while (cin >> s) {stack<char> opera;vector<int> numcnt;string s1;//后缀表达式//中缀表达式转后缀表达式for (int i = 0; i<s.size(); i++) {if (s[i] >= '0'&&s[i] <= '9') {int tmp = 0;while (s[i] >= '0'&&s[i] <= '9') {tmp++;s1 += s[i];i++;}i--;numcnt.push_back(tmp);}else if (s[i] == '-' || s[i] == '+') {if (s[i] == '-' && (s[i - 1] == '(' || s[i - 1] == '[' || s[i - 1] == '{'))s1 += '0';while (!opera.empty() && (opera.top() == '*' || opera.top() == '/' || opera.top() == '+' || opera.top() == '-')) {s1 += opera.top();opera.pop();}opera.push(s[i]);}else if (s[i] == '*' || s[i] == '/') {while (!opera.empty() && (opera.top() == '*' || opera.top() == '/')) {s1 += opera.top();opera.pop();}opera.push(s[i]);}else if (s[i] == '(' || s[i] == '[' || s[i] == '{')opera.push(s[i]);else if (s[i] == ')') {while (opera.top() != '(') {s1 += opera.top();opera.pop();}opera.pop();}else if (s[i] == ']') {while (opera.top() != '[') {s1 += opera.top();opera.pop();}opera.pop();}else if (s[i] == '}') {while (opera.top() != '{') {s1 += opera.top();opera.pop();}opera.pop();}elsecout << "Invalid input!" << endl;}while (!opera.empty()) {s1 += opera.top();opera.pop();}//计算后缀表达式的值stack<int> nums;int ind = 0;for (int i = 0; i<s1.size(); i++) {if (s1[i] >= '0'&&s1[i] <= '9') {int total = 0;while (numcnt[ind]--)total = 10 * total + (s1[i++] - '0');i--;nums.push(total);ind++;}else {int tmp1 = nums.top();nums.pop();int tmp2 = nums.top();nums.pop();if (s1[i] == '+')nums.push(tmp2 + tmp1);else if (s1[i] == '-')nums.push(tmp2 - tmp1);else if (s1[i] == '*')nums.push(tmp2*tmp1);elsenums.push(tmp2 / tmp1);}}cout << nums.top() << endl;}
}

 

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