【LeetCode】150. Evaluate Reverse Polish Notation 逆波兰表达式求值（Medium）（JAVA）
题目地址: https://leetcode.com/problems/evaluate-reverse-polish-notation/
题目描述：
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
题目大意
根据 逆波兰表示法，求表达式的值。
有效的运算符包括 +, -, *, / 。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。
解题方法
- 看例子的计算过程就可以明白：先把数组按序存到列表里；每当遇到符号，就把列表的后两个值拿出来计算，再把结果存到列表里
- note: 因为给定的表达式总是有效的，所以不需要判断是否有效
- 因为是后进先出，所以用 queue 比较快
class Solution {
public int evalRPN(String[] tokens) {
if (tokens.length <= 0) return 0;
LinkedList<Integer> list = new LinkedList<>();
for (int i = 0; i < tokens.length; i++) {
if ("+".equals(tokens[i]) || "-".equals(tokens[i]) || "*".equals(tokens[i]) || "/".equals(tokens[i])) {
int second = list.removeLast();
int first = list.removeLast();
int cur = 0;
if ("+".equals(tokens[i])) {
cur = first + second;
} else if ("-".equals(tokens[i])) {
cur = first - second;
} else if ("*".equals(tokens[i])) {
cur = first * second;
} else if ("/".equals(tokens[i])) {
cur = first / second;
}
list.add(cur);
} else {
list.add(Integer.parseInt(tokens[i]));
}
}
return list.get(0);
}
}
执行耗时:6 ms,击败了89.35% 的Java用户
内存消耗:38.1 MB,击败了90.03% 的Java用户