Merge pull request #1668 from axbybgl/axbybgl

Update 0150.逆波兰表达式求值.md
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程序员Carl
2022-10-02 10:47:03 +08:00
committed by GitHub

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@ -89,26 +89,29 @@ C++代码如下:
class Solution {
public:
int evalRPN(vector<string>& tokens) {
stack<int> st;
// 力扣修改了后台测试数据需要用longlong
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((long)num2 * (long)num1); //力扣改了后台测试数据
if (tokens[i] == "/") st.push(num2 / num1);
} else {
st.push(stoi(tokens[i]));
st.push(stoll(tokens[i]));
}
}
int 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;
}
};
```
## 题外话