LeetCode "Basic Calculator II"

Comparing with the 'I' one, there are two changes:

1. if last operator is "+" or "-", we don't evaluate expression
2. if last operator is "*" or "/", we evaluate it immediately, but we only pop out 2 operands.

class Solution {
struct Node
{
Node(int _v) : v(_v), isNum(true){}
Node(char b) : brack(b), isNum(false){}
bool isNum;
int v;
char brack;
};
public:
void singleEval(vector<Node> &operands, vector<char> &ops)
{
char c = ops.back(); ops.pop_back();
int v0 = operands.back().v; operands.pop_back();
int v1 = operands.back().v; operands.pop_back();
int ret;
if (c == '+') ret = v1 + v0;
if (c == '-') ret = v1 - v0;
if (c == '*') ret = v1 * v0;
if (c == '/') ret = v1 / v0;
operands.push_back(ret);
}
void eval(vector<Node> &operands, vector<char> &ops)
{
while (operands.size() > 1 && operands.back().isNum && operands[operands.size() - 2].isNum
&& ops.size() > 0)
{
singleEval(operands, ops);
}
}
int calculate(string s)
{
string ss;
for (auto c : s)
if (c != ' ') ss += c;

vector<Node> operands;
vector<char> ops;

size_t len = ss.length();
int i = 0;
int operand = 0;
while (i < len)
{
char c = ss[i];
if (isdigit(c))
{
operand *= 10;
operand += c - '0';
if (((i + 1) < len && !isdigit(ss[i + 1])) ||
(i + 1) == len)
{
operands.push_back(operand);
if (ops.size() > 0 && (ops.back() == '*' || ops.back() == '/'))
{
singleEval(operands, ops);
}
if (((i + 1) < len && ss[i + 1] != '*' && ss[i + 1] != '/')
|| (i + 1) == len)
eval(operands, ops);
operand = 0;
}
}
else if (c == '+' || c == '-' || c == '*' || c == '/')
{
ops.push_back(c);
}
else if (c == '(')
{
operands.push_back(Node('('));
}
else if (c == ')')
{
int v = operands.back().v; operands.pop_back();
char b = operands.back().brack; operands.pop_back();
operands.push_back(v);
eval(operands, ops);
}
i++;
}
eval(operands, ops);
return operands[0].v;
}
};