[LeetCode] 150. 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

逆波兰表达式，所有操作符置于操作数的后面，因此也被称为后缀表示法。逆波兰记法不需要括号来标识操作符的优先级。

逆波兰记法中，操作符置于操作数的后面。例如表达“三加四”时，写作“3 4 +”，而不是“3 + 4”。如果有多个操作符，操作符置于第二个操作数的后面，所以常规中缀记法的“3 - 4 + 5”在逆波兰记法中写作“3 4 - 5 +”：先3减去4，再加上5。使用逆波兰记法的一个好处是不需要使用括号。例如中缀记法中“3 - 4 * 5”与“（3 - 4）*5”不相同，但后缀记法中前者写做“3 4 5 * -”，无歧义地表示“3 (4 5 *) -”；后者写做“3 4 - 5 *”。

逆波兰表达式的解释器一般是基于堆栈的。解释过程一般是：操作数入栈；遇到操作符时，操作数出栈，求值，将结果入栈；当一遍后，栈顶就是表达式的值。因此逆波兰表达式的求值使用堆栈结构很容易实现，并且能很快求值。

注意：逆波兰记法并不是简单的波兰表达式的反转。因为对于不满足交换律的操作符，它的操作数写法仍然是常规顺序，如，波兰记法“/ 6 3”的逆波兰记法是“6 3 /”而不是“3 6 /”；数字的数位写法也是常规顺序。

解法1: 栈Stack

解法2: 递归

Java:

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public class Solution {     public int evalRPN(String[] tokens) {         int a,b;         Stack<Integer> S = new Stack<Integer>();         for (String s : tokens) {             if(s.equals("+")) {                 S.add(S.pop()+S.pop());             }             else if(s.equals("/")) {                 b = S.pop();                 a = S.pop();                 S.add(a / b);             }             else if(s.equals("*")) {                 S.add(S.pop() * S.pop());             }             else if(s.equals("-")) {                 b = S.pop();                 a = S.pop();                 S.add(a - b);             }             else {                 S.add(Integer.parseInt(s));             }         }           return S.pop();     } }

Java:

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public int evalRPN(String[] a) {   Stack<Integer> stack = new Stack<Integer>();       for (int i = 0; i < a.length; i++) {     switch (a[i]) {       case "+":         stack.push(stack.pop() + stack.pop());         break;                   case "-":         stack.push(-stack.pop() + stack.pop());         break;                   case "*":         stack.push(stack.pop() * stack.pop());         break;         case "/":         int n1 = stack.pop(), n2 = stack.pop();         stack.push(n2 / n1);         break;                   default:         stack.push(Integer.parseInt(a[i]));     }   }       return stack.pop(); }

Python:　　

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# Time:  O(n) # Space: O(n) import operator   class Solution:     # @param tokens, a list of string     # @return an integer     def evalRPN(self, tokens):         numerals, operators = [], {"+": operator.add, "-": operator.sub, "*": operator.mul, "/": operator.div}         for token in tokens:             if token not in operators:                 numerals.append(int(token))             else:                 y, x = numerals.pop(), numerals.pop()                 numerals.append(int(operators[token](x * 1.0, y)))         return numerals.pop()   if __name__ == "__main__":     print(Solution().evalRPN(["2", "1", "+", "3", "*"]))     print(Solution().evalRPN(["4", "13", "5", "/", "+"]))     print(Solution().evalRPN(["10","6","9","3","+","-11","*","/","*","17","+","5","+"]))

C++:

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

C++:　　

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class Solution { public:     int evalRPN(vector<string>& tokens) {         int op = tokens.size() - 1;         return helper(tokens, op);     }     int helper(vector<string>& tokens, int& op) {         string s = tokens[op];         if (s == "+" || s == "-" || s == "*" || s == "/") {             int v2 = helper(tokens, --op);             int v1 = helper(tokens, --op);             if (s == "+") return v1 + v2;             else if (s == "-") return v1 - v2;             else if (s == "*") return v1 * v2;             else return v1 / v2;         } else {             return stoi(s);         }     } };

　　

 

 

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