表达式求值

题目

#include <bits/stdc++.h>
using namespace std;
string a;
int shuzi(int l, int r)
{
	int sum = 0;
	for (int i = l; i <= r; i++)
	{
		sum = sum * 10 + a[i] - '0';
	}
	return sum;
}
int cal(int l, int r)
{

	int cnt = 0;
	int pos1 = -1, pos2 = -1, pos3 = -1;
	for (int i = l; i <= r; i++)
	{
		if (a[i] == '(')
			cnt++;
		if (a[i] == ')')
			cnt--;
		if (cnt <= 0)
		{ // 括号外或者没括号，定位最右边运算符位置（同优先级下，右边慢算）
			if (a[i] == '+' || a[i] == '-')
				pos1 = i;
			if (a[i] == '*' || a[i] == '/')
				pos2 = i;
			if (a[i] == '^')
				pos3 = i;
		}
	}
	if (pos1 == -1 && pos2 == -1 && pos3 == -1)
	{ // 括号内
		if (cnt > 0 && a[l] == '(')
			return cal(l + 1, r);
		if (cnt < 0 && a[r] == ')')
			return cal(l, r - 1);
		if (cnt == 0 && a[l] == '(')
			return cal(l + 1, r - 1); // 去括号
		return shuzi(l, r);
	}
	else if (pos1 != -1)
	{ // 把其他看作整体（即数学中的先算），然后把整体相加减
		if (a[pos1] == '+')
			return cal(l, pos1 - 1) + cal(pos1 + 1, r);
		else
			return cal(l, pos1 - 1) - cal(pos1 + 1, r);
	}
	else if (pos2 != -1)
	{
		if (a[pos2] == '*')
			return cal(l, pos2 - 1) * cal(pos2 + 1, r);
		else
			return cal(l, pos2 - 1) / cal(pos2 + 1, r);
	}
	else
	{
		return pow(cal(l, pos3 - 1), cal(pos3 + 1, r));
	}
}
int main()
{
	cin >> a;
	int len = a.size();
	cout << cal(0, len - 1);
}