[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。

 1 public class Solution {

 2 

 3     /** use interface as work-around for lambda expression */

 4     public interface Operator {

 5         int eval(int x, int y);

 6     }

 7 

 8     @SuppressWarnings("serial")

 9     public static final Map<String, Operator> OperatorMap = 

10     new HashMap<String, Solution.Operator>(){{

11         put("+", new Operator() {

12             public int eval(int x, int y) { return x + y; }});

13         put("-", new Operator() {

14             public int eval(int x, int y) { return x - y; }});

15         put("*", new Operator() {

16             public int eval(int x, int y) { return x * y; }});

17         put("/", new Operator() {

18             public int eval(int x, int y) { return x / y; }});

19     }};

20 

21     public int evalRPN(String[] tokens) {

22         Stack<Integer> operands = new Stack<Integer>();

23         for (String token : tokens) {

24             if (OperatorMap.containsKey(token)) {

25                 // for operator: calculate operation and then push to stack

26                 int op2 = operands.pop();

27                 int op1 = operands.pop();

28                 operands.push(OperatorMap.get(token).eval(op1, op2));

29             } else {

30                 // for operand: push to stack

31                 operands.push(Integer.parseInt(token));

32             }

33         }

34         return operands.pop();

35     }

36 

37 }