[Leetcode][150]Evaluate Reverse Polish Notation (Java)

题目在这里: https://leetcode.com/problems/evaluate-reverse-polish-notation/

【标签】Stack

【题目分析】思路就不多说了吧，大概就是，遇到数字这样的operand, 就先压到栈里面；遇到 +, -, *, / 的话，就可以进行局部的运算了，把栈顶的两个元素拿出来和operator进行运算，算好了，再压回栈里，留着以后继续运算。

【一点心得】 下面贴的是别人的好代码，特别值得一提的是，用了interface的方法，来模拟lambda表达式的功能，用function来作为parameter。而且也很好的增强了代码的可扩展性。

public class Solution {

    /** use interface as work-around for lambda expression */
    public interface Operator {
        int eval(int x, int y);
    }

    @SuppressWarnings("serial")
    public static final Map<String, Operator> OperatorMap = 
    new HashMap<String, Solution.Operator>(){{
        put("+", new Operator() {
            public int eval(int x, int y) { return x + y; }});
        put("-", new Operator() {
            public int eval(int x, int y) { return x - y; }});
        put("*", new Operator() {
            public int eval(int x, int y) { return x * y; }});
        put("/", new Operator() {
            public int eval(int x, int y) { return x / y; }});
    }};

    public int evalRPN(String[] tokens) {
        Stack<Integer> operands = new Stack<Integer>();
        for (String token : tokens) {
            if (OperatorMap.containsKey(token)) {
                // for operator: calculate operation and then push to stack
                int op2 = operands.pop();
                int op1 = operands.pop();
                operands.push(OperatorMap.get(token).eval(op1, op2));
            } else {
                // for operand: push to stack
                operands.push(Integer.parseInt(token));
            }
        }
        return operands.pop();
    }

}