Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /.
Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6之前已经讲过逆波兰表达式的生成和运算方法，实际上计算是更为简单的。具体见博客。
package com.liuhao.acm.leetcode;
import java.util.ArrayDeque;
import java.util.Deque;
/**
* 计算逆波兰表达式
*
* @author liuhao
*
* Evaluate the value of an arithmetic expression in Reverse Polish
* Notation.
*
* Valid operators are +, -, *, /. Each operand may be an integer or
* another expression.
*
* Some examples: ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
* ["4","13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
*
*/
public class EvalRPN {
public static int evalRPN(String[] tokens) {
// 栈，用于遍历初始字符串数组
Deque<Integer> stack = new ArrayDeque<Integer>();
int a, b;// 临时存放栈中弹出两个元素
/**
* 遍历初始字符串数组，
* 若当前字符为 运算符 ，则从栈中弹出两个元素，并用该运算符对它们进行运算，然后再将运算结果压入栈
* 若读到的是数字，则直接将其压入栈，不作其他操作
*/
for (int i = 0; i < tokens.length; i++) {
String temp = tokens[i];
switch (temp) {
case "+":
a = stack.pop();
b = stack.pop();
stack.push((b + a));
break;
case "-":
a = stack.pop();
b = stack.pop();
stack.push(b - a);
break;
case "*":
a = stack.pop();
b = stack.pop();
stack.push(b * a);
break;
case "/":
a = stack.pop();
b = stack.pop();
if (a == 0) {
return -1;
}
stack.push(b / a);
break;
default:
stack.push(Integer.parseInt(temp));
}
}
int result = stack.peek();
return result;
}
public static void main(String[] args) {
String[] input = new String[] { "4", "13", "5", "/", "+" };
System.out.println(evalRPN(input));
}
}
【LeetCode刷题Java版】Evaluate Reverse Polish Notation（计算逆波兰表达式）
原文地址：http://blog.csdn.net/bruce_6/article/details/40048643
