题目
- 请写一个整数计算器，支持加减乘三种运算和括号。
- 示例1
输入：“1+2”
返回值：3 - 示例2
输入：“(2*(3-4))*5”
返回值：-10 - 示例3
输入：“3+2*3*4-1”
返回值：26
思路
- 经典的中缀表达式求值。常用思路，将其转为后缀表达式（后缀表达式的特点：遇到运算符则将前两个数拿出来运算，便于计算机计算），边转边求值。
- 读到左括号 ( 较为特殊，读到左括号时直接入栈，
- 本题中运算符优先级：左括号或栈顶为空、乘、加减 、右括号。
- 读到右括号时一直出栈直到一个左括号出栈。
算法步骤
- 遍历字符串，当 遍历到的字符是数字 时，直接将其压入数字栈。
- 否则，当 遍历到的字符是运算符 时，判断字符，直到字符串读完：
- 若符号栈为空，则直接将其压入符号栈。
- 若符号栈非空，则判断符号栈栈顶的运算符优先级：
- 若 遍历到的运算符符号的优先级 > 符号栈栈顶元素的优先级，或符号栈空，则将读取的运算符压入符号栈。
- 若 遍历到的运算符符号的优先级 ≤ 符号栈栈顶元素的优先级，则将其出栈，并从数字栈弹出两个元素进行该运算，先弹出的为右操作数，后弹出的为左操作数，将结果压入数字栈。
- 最后，若符号栈非空，则弹出栈顶元素，并从数字栈弹出两个元素来进行该运算，重复该操作直到符号栈空。
- 符号栈为空后，弹出数字栈栈顶元素作为结果。
- 返回数值站栈顶元素作为最终结果。
int solve(string s) {
stack<char> sign;
stack<int> num;
for (int i = 0; i < s.length(); i++)
{
if (s[i] >= '0' && s[i] <= '9')
{
int rear = i;
while ((rear + 1) < s.length() && s[rear + 1] >= '0' && s[rear + 1] <= '9')
{
rear++;
}
int n = 0;
for (int j = rear; j >= i; j--) {
n += (s[j] - '0') * pow(10, (rear - j));
}
i = rear;
num.push(n);
}
else if (s[i] == '(' || s[i] == '*'|| sign.size() == 0 )
{
sign.push(s[i]);
}
else if (s[i] == '+' || s[i] == '-')
{
while (1)
{
if (sign.size() == 0)
break;
if (sign.top() == '*')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 * n2);
}
else if (sign.top() == '+')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 + n2);
}
else if (sign.top() == '-')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 - n2);
}
else {
break;
}
}
sign.push(s[i]);
}
else if (s[i] == ')')
{
while (1)
{
if (sign.size() == 0)
{
break;
}
else if (sign.top() == '(')
{
sign.pop();
break;
}
else if (sign.top() == '+')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 + n2);
}
else if (sign.top() == '-')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 - n2);
}
else if (sign.top() == '*')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 * n2);
}
}
}
}
while (sign.size() != 0)
{
if (sign.top() == '+')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 + n2);
}
else if (sign.top() == '-')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 - n2);
}
else if (sign.top() == '*')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 * n2);
}
}
return num.top();
}
实例
#include <iostream>
#include <stack>
using namespace std;
int solve(string s) {
stack<char> sign;
stack<int> num;
for (int i = 0; i < s.length(); i++)
{
if (s[i] >= '0' && s[i] <= '9')
{
int rear = i;
while ((rear + 1) < s.length() && s[rear + 1] >= '0' && s[rear + 1] <= '9')
{
rear++;
}
int n = 0;
for (int j = rear; j >= i; j--) {
n += (s[j] - '0') * pow(10, (rear - j));
}
i = rear;
num.push(n);
}
else if (s[i] == '(' || s[i] == '*'|| sign.size() == 0 )
{
sign.push(s[i]);
}
else if (s[i] == '+' || s[i] == '-')
{
while (1)
{
if (sign.size() == 0)
break;
if (sign.top() == '*')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 * n2);
}
else if (sign.top() == '+')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 + n2);
}
else if (sign.top() == '-')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 - n2);
}
else {
break;
}
}
sign.push(s[i]);
}
else if (s[i] == ')')
{
while (1)
{
if (sign.size() == 0)
{
break;
}
else if (sign.top() == '(')
{
sign.pop();
break;
}
else if (sign.top() == '+')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 + n2);
}
else if (sign.top() == '-')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 - n2);
}
else if (sign.top() == '*')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 * n2);
}
}
}
}
while (sign.size() != 0)
{
if (sign.top() == '+')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 + n2);
}
else if (sign.top() == '-')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 - n2);
}
else if (sign.top() == '*')
{
sign.pop();
int n2 = num.top();
num.pop();
int n1 = num.top();
num.pop();
num.push(n1 * n2);
}
}
return num.top();
}
int main()
{
string mys = "1+3*(5+2)";
cout<< mys<<"=" << solve(mys) << endl;
return 0;
}
