用c++语言中的栈实现简单表达式求值并给出代码
时间: 2024-10-15 14:22:57 浏览: 35
在 C++ 中,我们可以使用栈(Stack)来实现简单的表达式求值,例如逆波兰表示法(Reverse Polish Notation,RPN)。这种方法也被称为后缀表示法,因为操作数在操作符之后。这样可以避免括号的复杂处理。
首先,我们需要定义两个栈:一个用于存储操作数,另一个用于临时存放操作符。然后遍历输入的表达式,根据遇到的字符(数字或运算符)来执行相应的操作:
1. 对于数字,直接压入操作数栈。
2. 遇到运算符时,从操作数栈弹出两个操作数,并对它们进行运算(如加减乘除),将结果压回操作数栈,同时移动运算符到临时栈。
3. 当遍历完所有字符后,如果还有运算符留在临时栈里,需要继续进行运算。
下面是一个简单的 C++ 示例代码实现:
```cpp
#include <stack>
#include <string>
int eval(std::string tokens) {
std::stack<int> numStack;
std::stack<char> opStack;
for (char token : tokens) {
if (isdigit(token)) {
int num = 0;
while (isdigit(token)) {
num = num * 10 + token - '0';
++token;
}
numStack.push(num);
} else if (token == '+' || token == '-' || token == '*' || token == '/') {
while (!opStack.empty() && hasPrecedence(token, opStack.top())) {
int right = numStack.top();
numStack.pop();
int left = numStack.top();
numStack.pop();
char op = opStack.top();
opStack.pop();
switch (op) {
case '+':
numStack.push(left + right);
break;
case '-':
numStack.push(left - right);
break;
case '*':
numStack.push(left * right);
break;
case '/':
if (right != 0)
numStack.push(left / right);
else
throw std::runtime_error("Division by zero");
break;
}
}
opStack.push(token);
}
}
// Handle any remaining operators
while (!opStack.empty()) {
int right = numStack.top();
numStack.pop();
int left = numStack.top();
numStack.pop();
char op = opStack.top();
opStack.pop();
switch (op) {
case '+':
numStack.push(left + right);
break;
case '-':
numStack.push(left - right);
break;
case '*':
numStack.push(left * right);
break;
case '/':
numStack.push(left / right);
break;
}
}
return numStack.top(); // Return the final result
}
bool hasPrecedence(char op1, char op2) {
if (op2 == '(' || op2 == ')') return false; // Ignore parentheses
if ((op1 == '+' || op1 == '-') && (op2 == '*' || op2 == '/')) return true;
return op1 > op2;
}
// Usage example
std::string expression = "2 3 + 4 *";
int result = eval(expression);
std::cout << "Result: " << result << std::endl;
```
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