[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:
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:
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:
# 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++:
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++:
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);
}
}
};
