简单的计算器
题目:
设计一个计算器,输入一个字符串存储的数学表达式,可以计算包括 “(“、
”)”、”+”、”-”四种符号的数学表达式,输入的数学表达式字符串保证是合
法的。输入的数学表达式中可能存在空格字符。
如计算:
“(1 + 1)" = 2
"1+121 - (14+ (5-6) ) = 109
PS:这题一点也不简单。
解题图解:
解题代码:
#include <iostream>
#include<string>
#include<stack>
using namespace std;
void compute(stack<int> &number_stack,
stack<char> &operation_stack) {
if (number_stack.size() < 2)
return;
int num2 = number_stack.top();
number_stack.pop();
int num1 = number_stack.top();
number_stack.pop();
if (operation_stack.top() == '+') {
number_stack.push(num1 + num2);
}
else if (operation_stack.top() == '-') {
number_stack.push(num1 - num2);
}
operation_stack.pop();
}
class Calculator
{
public:
int calculator(string s) {
//定义三个常量记录这三个点
static const int STATE_BEGIN = 0;
static const int NUMBER_STATE = 1;
static const int OPERATION_STATE = 2;
stack<int> number_stack;
stack<char> operation_stack;
int number = 0;
int STATE = STATE_BEGIN;
int computer_flag = 0;
for (int i = 0; i < s.length(); i++)
{
if (s[i] == ' ') {
continue;
}
switch (STATE)
{
case STATE_BEGIN:
if (s[i] >= '0' && s[i <= '9'])
{
STATE = NUMBER_STATE;
}
else {
STATE = OPERATION_STATE;
}
i--;
break;
case NUMBER_STATE:
if (s[i] >= '0' && s[i <= '9'])
{
number = number * 10 + s[i] - '0';
}
else {
number_stack.push(number);
if (computer_flag == 1) {
compute(number_stack, operation_stack);
}
number = 0;
i--;
STATE = OPERATION_STATE;
}
break;
case OPERATION_STATE:
if (s[i] == '+' || s[i] == '-')
{
operation_stack.push(s[i]);
computer_flag = 1;
}
else if (s[i] == '(')
{
STATE = NUMBER_STATE;
computer_flag = 0;
}
else if (s[i] >= '0' && s[i] <= '9') {
STATE = NUMBER_STATE;
i--;
}
else if (s[i == ')']) {
compute(number_stack, operation_stack);
}
break;
}
}
if (number != 0)
{
number_stack.push(number);
compute(number_stack, operation_stack);
}
if (number == 0 && number_stack.empty())
{
return 0;
}
return number_stack.top();
}
};
int main()
{
//测试
string s = "1+121 - (14+ (5-6) )";
Calculator calculator;
int result = calculator.calculator(s);
cout << "结果是: " << result << endl;
}