[LeetCode] 150. Evaluate Reverse Polish Notation

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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。题意是根据逆波兰表达式的规则求值，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。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public int evalRPN(String[] tokens) {
 3         Stack<Integer> stack = new Stack<>();
 4         for (String s : tokens) {
 5             if (s.equals("+")) {
 6                 stack.push(stack.pop() + stack.pop());
 7             } else if (s.equals("-")) {
 8                 int a = stack.pop();
 9                 int b = stack.pop();
10                 stack.push(b - a);
11             } else if (s.equals("*")) {
12                 stack.push(stack.pop() * stack.pop());
13             } else if (s.equals("/")) {
14                 int a = stack.pop();
15                 int b = stack.pop();
16                 stack.push(b / a);
17             } else {
18                 stack.push(Integer.parseInt(s));
19             }
20         }
21         return stack.pop();
22     }
23 }

LeetCode 题目总结

[LeetCode] 150. Evaluate Reverse Polish Notation

标签：als   new   etc   遇到   The   and   inpu   war   run   

原文地址：https://www.cnblogs.com/cnoodle/p/12899749.html