LeetCode Evaluate Reverse Polish Notation

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are

+

,

-

,

*

,

/

. Each operand may be an integer or another expression.

Some examples:

["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

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class Solution {
public:
int evalRPN(vector<string> &tokens) {
int len = tokens.size();
if (len < 1) return 0;

vector<int> stack;
for (int i=0; i<len; i++) {
string &p = tokens[i];
if (p[0] >= '0' && p[0] <= '9' || p[0] == '-' && p.size() > 1) {
stack.push_back(stoi(p));
} else {
int slen = stack.size();
stack[slen - 2] = op(p[0], stack[slen - 2], stack[slen - 1]);
stack.pop_back();
}
}

return stack[0];
}

int op(const char op, int a, int b) {
int res = 0;
switch(op) {
case '+': res = a + b; break;
case '-': res = a - b; break;
case '/': res = a / b; break;
case '*': res = a * b; break;
}
return res;
}

int stoi(string &s) {
int len = s.size();
if (len < 1) return 0;
bool neg = s[0] == '-';
int i = neg ? 1 : 0;
int res = 0;
while (i < len) {
res = res * 10 + s[i] - '0';
i++;
}
return neg ? -res : res;
}
};

数据结构栈基础，一次过

时隔一年智商又下降了：

class Solution {
public:
int evalRPN(vector<string>& tokens) {
stack<int> nums;
for (string& token : tokens) {
if (!isop(token)) {
int d = 0;
sscanf(token.c_str(), "%d", &d);
nums.push(d);
} else {
int b = nums.top();
nums.pop();
int a = nums.top();
nums.pop();
nums.push(calc(a, b, token[0]));
}
}
return nums.top();
}
bool isop(string& str) {
return str.size() == 1 && (str[0] == '+' || str[0] == '-' || str[0] == '*' || str[0] == '/');
}

int calc(int a, int b, char op) {
switch (op) {
case '+': return a + b; break;
case '-': return a - b; break;
case '/': return a / b; break;
case '*': return a * b; break;
}
return 0;
}
};