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224+Basic Calculator.cpp
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224+Basic Calculator.cpp
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class Solution {
public:
unordered_map<char, int> ump = {
{'-', 1},
{'+', 1},
{'*', 2},
{'/', 2},
{'%', 2},
{'^', 3},
};
stack<long long> nums;
stack<char> ops;
void eval() {
long long b = nums.top(); nums.pop();
long long a = nums.top(); nums.pop();
char op = ops.top(); ops.pop();
long long res = 0;
switch (op) {
case '+': res = a + b; break;
case '-': res = a - b; break;
case '*': res = a * b; break;
case '/': res = a / b; break;
case '%': res = a % b; break;
case '^': res = pow(a, b); break;
}
nums.push(res);
}
int calculate(string s) {
nums.push(0);
for (int i = 0; i < s.size(); ++i) {
if (s[i] == ' ') continue;
if (isdigit(s[i])) { // 数字
int res = 0;
while (i < s.size() && isdigit(s[i])) {
res = res * 10 + (s[i++] - '0');
}
nums.push(res);
i--;
} else { // 运算符
if (s[i] == '(') {
ops.push(s[i]);
if (i + 1 < s.size() && s[i + 1] == '-') nums.push(0);
} else if (s[i] == ')') { // 计算该括号内内容
while (ops.top() != '(') { // 找到前一个 ( 并算其中的值
eval();
}
ops.pop();
} else {
char nowOp = s[i];
// 有一个新操作要入栈时,先把栈内可以算的都算了
// 但注意的是 只有满足「栈内运算符」比「当前运算符」优先级高/同等,才进行运算
while (!ops.empty() && ops.top() != '(' && ump[ops.top()] >= ump[nowOp]) {
eval();
}
ops.push(nowOp);
}
}
}
while(!ops.empty() && ops.top() != '(') {
eval();
}
return nums.top();
}
};