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功能说明
- 实现+、-、*、/、括号的整数运算
- 实现处理多余空格
实现1（更高效）
- 用引用
l指针的方式逐步处理 1
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| int core(string& s, int& l) {
stack<int> stk;
int n = s.size();
int num = 0;
char sign = '+';
for (; l < n; ++l) {
char c = s[l];
if (isdigit(c))
num = num * 10 + (c - '0');
if ((!isdigit(c) && c != ' ') || l == n - 1) {
if (c == '(')
num = core(s, ++l);
int prev;
switch (sign) {
case '+':
stk.push(num); break;
case '-':
stk.push(-num); break;
case '*':
prev = stk.top(); stk.pop();
stk.push(prev * num);
break;
case '/':
prev = stk.top(); stk.pop();
stk.push(prev / num);
break;
default: break;
}
sign = c;
num = 0;
if (c == ')')
break;
}
}
int ret = 0;
while (!stk.empty()) {
ret += stk.top();
stk.pop();
}
return ret;
}
int caculator(string s) {
int tmp = 0;
return core(s, tmp);
}
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实现2（更优雅）
- 由于需要不停
pop首部，所以采用deque代替移动的l，更优雅 1
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| int core(deque<char>& s) {
stack<int> stk;
int num = 0;
char sign = '+';
while (!s.empty()) {
char c = s[0]; s.pop_front();
if (isdigit(c))
num = num * 10 + (c - '0');
if ((!isdigit(c) && c != ' ') || s.empty()) {
if (c == '(')
num = core(s);
int prev;
switch (sign) {
case '+':
stk.push(num); break;
case '-':
stk.push(-num); break;
case '*':
prev = stk.top(); stk.pop();
stk.push(prev * num);
break;
case '/':
prev = stk.top(); stk.pop();
stk.push(prev / num);
break;
default: break;
}
sign = c;
num = 0;
if (c == ')')
break;
}
}
int ret = 0;
while (!stk.empty()) {
ret += stk.top();
stk.pop();
}
return ret;
}
int caculator(string s) {
deque<char> dq;
for (auto& e : s) dq.push_back(e);
return core(dq);
}
|
参考