[leecode]Evaluate Reverse Polish Notation

Evaluate Reverse Polish Notation

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 }

第二遍：

有if-else语句来代替switch，基本思想一样

 1 public class Solution {
 2     public int evalRPN(String[] tokens) {
 3         if(tokens == null || tokens.length == 0) return 0;
 4         LinkedList<Integer> stack = new LinkedList<Integer>();
 5         for(int i = 0; i < tokens.length; i++){
 6             if("+".equals(tokens[i])){
 7                 int tem = stack.pop();
 8                 stack.push(stack.pop() + tem);
 9             }else if("-".equals(tokens[i])){
10                 int tem = stack.pop();
11                 stack.push(stack.pop() - tem);
12             }else if("*".equals(tokens[i])){
13                 int tem = stack.pop();
14                 stack.push(stack.pop() * tem);
15             }else if("/".equals(tokens[i])){
16                 int tem = stack.pop();
17                 stack.push(stack.pop() / tem);
18             }else{
19                 stack.push(Integer.valueOf(tokens[i]));
20             }
21         }
22         return stack.pop();
23     }
24 }

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