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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.Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
算法思路：
很经典的stack应用，书中例题，遇到数字压栈，遇到符号弹栈，并将结果压栈，最后返回栈顶元素。
1 public class Solution { 2 public int evalRPN(String[] tokens) { 3 if(tokens == null || tokens.length == 0) return 0; 4 String[] operand = new String[]{"+","-","*","/"}; 5 Set<String> set = new HashSet<String>(Arrays.asList(operand)); 6 Stack<Integer> num = new Stack<Integer>(); 7 for(String s : tokens){ 8 if(!set.contains(s)){ 9 num.push(Integer.valueOf(s)); 10 }else{ 11 int b = num.pop(); 12 int a = num.pop(); 13 switch(s){ 14 case "*": num.push(a * b); break; 15 case "+": num.push(a + b); break; 16 case "-": num.push(a - b); break; 17 case "/": num.push(a / b); break; 18 default : break; 19 } 20 } 21 } 22 return num.pop(); 23 } 24 }
[leecode]Evaluate Reverse Polish Notation,布布扣,bubuko.com
[leecode]Evaluate Reverse Polish Notation
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原文地址：http://www.cnblogs.com/huntfor/p/3891676.html
