Update 0150.逆波兰表达式求值.md

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axbybgl
2022-09-29 10:51:23 +08:00
parent fd820f6f4f
commit f35ecefd54

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@ -89,26 +89,32 @@ C++代码如下:
class Solution {
public:
int evalRPN(vector<string>& tokens) {
stack<int> st;
// 考虑到第20个样例使用int会溢出
// 此处使用long long来存储number
// 在最后用int()强行转换成int输出
stack<long long> st;
for (int i = 0; i < tokens.size(); i++) {
if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
int num1 = st.top();
long long num1 = st.top();
st.pop();
int num2 = st.top();
long long num2 = st.top();
st.pop();
if (tokens[i] == "+") st.push(num2 + num1);
if (tokens[i] == "-") st.push(num2 - num1);
if (tokens[i] == "*") st.push(num2 * num1);
if (tokens[i] == "/") st.push(num2 / num1);
} else {
st.push(stoi(tokens[i]));
st.push(atoll(tokens[i].c_str()));
}
}
int result = st.top();
long long result = st.top();
st.pop(); // 把栈里最后一个元素弹出(其实不弹出也没事)
return result;
return int(result);
}
};
```
## 题外话