150. Evaluate Reverse Polish Notation (M)

Evaluate Reverse Polish Notation (M)

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
           

题意

计算逆波兰表达式(即后缀表达式)的值。

思路

遇到数字就压栈；遇到符号则从栈中弹出两个数字进行运算(注意后出栈的在前，先出栈的在后)，将得到的结果再压入栈中；最后栈中只剩一个数，即所求结果。

代码实现

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> stack = new ArrayDeque<>();

        for (String token : tokens) {
            if (token.equals("+")) {
                int y = stack.pop(), x = stack.pop();
                stack.push(x + y);
            } else if (token.equals("-")) {
                int y = stack.pop(), x = stack.pop();
                stack.push(x - y);
            } else if (token.equals("*")) {
                int y = stack.pop(), x = stack.pop();
                stack.push(x * y);
            } else if (token.equals("/")) {
                int y = stack.pop(), x = stack.pop();
                stack.push(x / y);
            } else {
                stack.push(Integer.parseInt(token));
            }
        }

        return stack.pop();
    }
}