[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。

public class Solution {
    public int evalRPN(String[] tokens) {
        if(tokens == null || tokens.length == 0) return 0;
        String[] operand = new String[]{"+","-","*","/"};
        Set<String> set = new HashSet<String>(Arrays.asList(operand));
        Stack<Integer> num = new Stack<Integer>();
        for(String s : tokens){
            if(!set.contains(s)){
                num.push(Integer.valueOf(s));
            }else{
                int b = num.pop();
                int a = num.pop();
                switch(s){
                    case "*": num.push(a * b); break;
                    case "+": num.push(a + b); break;
                    case "-": num.push(a - b); break;
                    case "/": num.push(a / b); break;
                    default : break;
                }
            }
        }
        return num.pop();
    }
}

 

第二遍：

有if-else语句来代替switch，基本思想一样

public class Solution {
    public int evalRPN(String[] tokens) {
        if(tokens == null || tokens.length == 0) return 0;
        LinkedList<Integer> stack = new LinkedList<Integer>();
        for(int i = 0; i < tokens.length; i++){
            if("+".equals(tokens[i])){
                int tem = stack.pop();
                stack.push(stack.pop() + tem);
            }else if("-".equals(tokens[i])){
                int tem = stack.pop();
                stack.push(stack.pop() - tem);
            }else if("*".equals(tokens[i])){
                int tem = stack.pop();
                stack.push(stack.pop() * tem);
            }else if("/".equals(tokens[i])){
                int tem = stack.pop();
                stack.push(stack.pop() / tem);
            }else{
                stack.push(Integer.valueOf(tokens[i]));
            }
        }
        return stack.pop();
    }
}