Update 0150.逆波兰表达式求值.md

力扣修改了后端测试数据
This commit is contained in:
程序员Carl
2022-10-02 10:46:19 +08:00
committed by GitHub
parent f35ecefd54
commit b0ec03922e

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@ -89,10 +89,7 @@ C++代码如下:
class Solution { class Solution {
public: public:
int evalRPN(vector<string>& tokens) { int evalRPN(vector<string>& tokens) {
// 考虑到第20个样例使用int会溢出 // 力扣修改了后台测试数据需要用longlong
// 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; stack<long long> st;
for (int i = 0; i < tokens.size(); i++) { for (int i = 0; i < tokens.size(); i++) {
if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") { if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
@ -105,13 +102,13 @@ public:
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 { } else {
st.push(atoll(tokens[i].c_str())); st.push(stoll(tokens[i]));
} }
} }
long long result = st.top(); 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) st.pop(); // 把栈里最后一个元素弹出(其实不弹出也没事)
return int(result); return result;
} }
}; };