227. Basic Calculator II**
https://leetcode.com/problems/basic-calculator-ii/
题目描述
Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.
Example 1:
Input: "3+2*2"
Output: 7
Example 2:
Input: " 3/2 "
Output: 1
Example 3:
Input: " 3+5 / 2 "
Output: 5
Note:
- You may assume that the given expression is always valid.
- Do not use the eval built-in library function.
C++ 实现 1
原来写计算器也不是那么容易的…
完成本题参考了:
我最后的解法和他们的不同, 原因在于我想按照自己的方式实现一下. 最后发现实现的代码和他们的不同. C++ 实现 1 先介绍我的思路.
class Solution {
public:
int calculate(string s) {
// 使用 op 记录当前遇到的符号, 用 prev 记录上一个数, 用 num 记录当前遇到的数
s = s + "++";
long long prev = 0, num = 0, total = 0;
char op = '\0';
for (int i = 0; i < s.size(); ++ i) {
// cout << num << endl;
if (std::isdigit(s[i])) num = num * 10 + s[i] - '0';
else if (!std::isdigit(s[i]) && !std::isspace(s[i])) {
if (op == '\0') prev = num;
if (op == '+' || op == '-') {
total += prev;
num *= op == '-' ? -1 : 1;
prev = num;
} else if (op == '*') prev *= num;
else if (op == '/') prev /= num;
op = s[i];
num = 0;
}
}
return total;
}
};
C++ 实现 2
来自: 17 lines C++, easy, 20 ms, 我做了一点修改. 代码很好理解, 不多介绍.
class Solution {
public:
int calculate(string s) {
istringstream in('+' + s + '+');
long long total = 0, term = 0, n;
char op;
while (in >> op) {
if (op == '+' or op == '-') {
total += term;
in >> term;
term *= op == '-' ? -1 : 1;
} else {
in >> n;
if (op == '*')
term *= n;
else
term /= n;
}
}
return total;
}
};
C++ 实现 3
参考: Share my java solution, 修改为 C++;
class Solution {
public:
int calculate(string s) {
stack<long long> st;
long long num = 0;
char sign = '+';
for (int i = 0; i < s.size(); ++ i) {
if (std::isdigit(s[i])) num = num * 10 + s[i] - '0';
if ((!std::isdigit(s[i]) && !std::isspace(s[i])) || (i == s.size() - 1)) {
if (sign == '-') st.push(-num);
if (sign=='+') st.push(num);
if (sign=='*'){
auto val = st.top();
st.pop();
st.push(val * num);
}
if (sign=='/'){
auto val = st.top();
st.pop();
st.push(val / num);
}
sign = s[i];
num = 0;
}
}
long long res = 0;
while (!st.empty()) {
res += st.top();
st.pop();
}
return res;
}
};
