[LeetCode] 150. Evaluate Reverse Polish Notation

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

逆波兰表达式求值。

根据 逆波兰表示法，求表达式的值。

有效的算符包括 +、-、*、/ 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。

说明：

整数除法只保留整数部分。
给定逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/evaluate-reverse-polish-notation
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思路是用stack。注意到这里的所有计算并不遵循乘除法优先于加减法的原则，而是先遇到什么符号就立马进行计算。思路直接参见代码。

时间O(n)

空间O(n)

Java实现

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

LeetCode 题目总结