[LeetCode] Evaluate Reverse Polish Notation 计算逆波兰表达式

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are `+``-``*``/`. Each operand may be an integer or another expression.

Note:

• Division between two integers should truncate toward zero.
• The given RPN expression is always valid. That means the expression would always evaluate to a result and there won’t be any divide by zero operation.

Example 1:

```Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9
```

Example 2:

```Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
```

Example 3:

```Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22```

```class Solution {
public:
int evalRPN(vector<string>& tokens) {
if (tokens.size() == 1) return stoi(tokens[0]);
stack<int> st;
for (int i = 0; i < tokens.size(); ++i) {
if (tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/") {
st.push(stoi(tokens[i]));
} else {
int num1 = st.top(); st.pop();
int 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);
}
}
return st.top();
}
};```

```class Solution {
public:
int evalRPN(vector<string>& tokens) {
int op = (int)tokens.size() - 1;
return helper(tokens, op);
}
int helper(vector<string>& tokens, int& op) {
string str = tokens[op];
if (str != "+" && str != "-" && str != "*" && str != "/") return stoi(str);
int num1 = helper(tokens, --op);
int num2 = helper(tokens, --op);
if (str == "+") return num2 + num1;
if (str == "-") return num2 - num1;
if (str == "*") return num2 * num1;
return num2 / num1;
}
};```

Basic Calculator

https://leetcode.com/problemset/algorithms/

https://leetcode.com/problems/evaluate-reverse-polish-notation/discuss/47642/a-recursive-solution-in-cpp

https://leetcode.com/problems/evaluate-reverse-polish-notation/discuss/47544/Challenge-me-neat-C%2B%2B-solution-could-be-simpler

原文作者：Grandyang
原文地址: http://www.cnblogs.com/grandyang/p/4247718.html
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