題目描述

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

我的解題思路：這道題應該是簡單的，但是要註意Interge.parseInt()方法以及棧中數據的先後順序

在下的代碼 時間122ms 空間11880k

import java.util.Stack;

public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> stack = new Stack<Integer>();
        for (int i=0;i<tokens.length;i++){
            String str = tokens[i];
            if (str.equals("+")) {
                int res = stack.pop() + stack.pop();
                stack.push(res);
            }
            else if (str.equals("-")) {
                int res = (stack.pop() - stack.pop()) * (-1);
                stack.push(res);
            }
            else if (str.equals("*")) {
                int res = stack.pop() * stack.pop();
                stack.push(res);
            }
            else if (str.equals("/")) {
                int num1 = stack.pop();
                int num2 = stack.pop();
                int res = num2 / num1;
                stack.push(res);
            }
            else {
                stack.push(Integer.parseInt(str));
            }
        }
      	return stack.pop();
    }
}

大神代碼 時間148ms 空間13132k 思路很新穎，用了異常來pop元素，非常好的方法

import java.util.Stack;
public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> stack = new Stack<Integer>();
        for(int i = 0;i<tokens.length;i++){
            try{
                int num = Integer.parseInt(tokens[i]);
                stack.add(num);
            }catch (Exception e) {
                int b = stack.pop();
                int a = stack.pop();
                stack.add(get(a, b, tokens[i]));
            }
        }
        return stack.pop();
    }
    private int get(int a,int b,String operator){
        switch (operator) {
        case "+":
            return a+b;
        case "-":
            return a-b;
        case "*":
            return a*b;
        case "/":
            return a/b;
        default:
            return 0;
        }
    }
}

leetcode evaluate-reverse-polish-notation