波兰式解析Evaluate Reverse Polish Notation

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本文介绍波兰式解析算法。

Leetcode 原题如下

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

中文翻译如下，有效的操作符包括： +， -， *，/。每个操作数是一个整数或者是另个一表达式。

例子如上图。

Java实现如下，包括两个版本。

版本一，包括打印解析树，及"((2+1)*3)->".

import java.util.Stack;public class Solution {public static void main(String[] args){String[] input1 = {"2", "1", "+", "3", "*"};String[] input2 = {"4", "13", "5", "/", "+"};Solution s = new Solution();s.evalRPN(input1);s.evalRPN(input2);}    class Node{        int val;        String sVal;        Node(int val, String sval){            this.val = val;            this.sVal = sval;        }    }    private boolean isNumber(String str){        return str.matches("-?\\d+");    }        private boolean isOperator(String str){        return str.matches("[/+*-]");    }        private int compute(int left, int right, char operator){        switch(operator){            case '/': return left / right;            case '*': return left * right;            case '-': return left - right;            case '+': return left + right;            //case default return 0;//need to handle this one        }return 0;    }    public int evalRPN(String[] tokens) {        Stack<Node> stack = new Stack();        for(int i = 0; i < tokens.length; i++){            String token = tokens[i];            if(isNumber(token)) {                int val = Integer.parseInt(token);                stack.push(new Node(val, token));            }else if(isOperator(token)){                Node right = stack.pop();                Node left = stack.pop();                int tmp = compute(left.val, right.val, token.charAt(0));                String stmp = "("+left.sVal + token + right.sVal+")";                Node rst = new Node(tmp, stmp);                stack.push(rst);            }        }        int rval = 0;        if(stack.size() == 1){            Node node = stack.pop();            System.out.println(node.sVal + "->" + node.val);            rval = node.val;        }        return rval;    }}

输出结果如下：

((2+1)*3)->9

(4+(13/5))->6

版本二，简化版本，速度更快

public class Solution {    private boolean isNumber(String str){        return str.matches("-?\\d+");    }        private boolean isOperator(String str){        return str.matches("[/+*-]");    }        private int compute(int left, int right, char operator){        switch(operator){            case '/': return left / right;            case '*': return left * right;            case '-': return left - right;            case '+': return left + right;            //case default return 0;//need to handle this one        }return 0;    }    public int evalRPN(String[] tokens) {        Stack<Integer> stack = new Stack();        for(int i = 0; i < tokens.length; i++){            String token = tokens[i];            if(isNumber(token)) {                int val = Integer.parseInt(token);                stack.push(val);            }else if(isOperator(token)){                int right = stack.pop();                int left = stack.pop();                int tmp = compute(left,right, token.charAt(0));                stack.push(tmp);            }        }        int rval = 0;        if(stack.size() == 1){            rval = stack.pop();        }        return rval;    }}

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