leetcode150. Evaluate Reverse Polish Notation

题目要求

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

计算后缀表达式。我们一般看到的数学表达式就是中缀表达式，也就是将符号放在两个数字之间。后缀表达式也就是将运算符放在相应数字的后面。后缀表达式相当于树中的后序遍历。

思路一：栈

当我们遇到数字时就将数字压入栈中，如果遇到操作符就将栈顶的两个数字弹出，并将其根据操作符计算结构并重新压入栈中。栈中剩下的最后的值就是我们的结果。

    public int evalRPN(String[] tokens) {
        LinkedList<Integer> stack = new LinkedList<Integer>();
        for(String token : tokens){
            if(token.equals("+")){
                int operand1 = stack.pop();
                int operand2 = stack.pop();
                stack.push(operand2 + operand1);
            }else if(token.equals("-")){
                int operand1 = stack.pop();
                int operand2 = stack.pop();
                stack.push(operand2 - operand1);
            }else if(token.equals("*")){
                int operand1 = stack.pop();
                int operand2 = stack.pop();
                stack.push(operand2 * operand1);
            }else if(token.equals("/")){
                int operand1 = stack.pop();
                int operand2 = stack.pop();
                stack.push(operand2 / operand1);
            }else{
                stack.push(Integer.valueOf(token));
            }
        }
        return stack.pop();
    }

思路二：递归

从后缀表达式的末尾开始递归获取操作符对应的两个操作符。通过index获得对应位置的操作符。如果对应的还是操作符，则继续递归往前计算。

    int index;
    public int evalRPN2(String[] tokens){
        index = tokens.length-1;
        return recursive(tokens);
    } 
    public int recursive(String[] tokens){
        String current = tokens[index--];
        int operand1, operand2;
        switch(current){
        case "+" : 
            operand1 = recursive(tokens);
            operand2 = recursive(tokens);
            return operand1 + operand2;
        case "-" :
            operand1 = recursive(tokens);
            operand2 = recursive(tokens);
            return operand2 - operand1;
        case "*" :
            operand1 = recursive(tokens);
            operand2 = recursive(tokens);
            return operand2 * operand1;
        case "/" :
            operand1 = recursive(tokens);
            operand2 = recursive(tokens);
            return operand2 / operand1;
        default:
            return Integer.valueOf(current);
        }
    }

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