LeetCode|Evaluate Reverse Polish Notation-优快云博客

本文链接：https://blog.youkuaiyun.com/hwb1992/article/details/24197851

本文介绍了一种使用栈的数据结构解决逆波兰表达式求值问题的方法。通过遍历输入的表达式，当遇到操作数时将其压入栈中，当遇到运算符时则从栈中弹出两个操作数进行计算并将结果重新压入栈中，最终栈顶元素即为表达式的计算结果。

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题目

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

思路

活活一个水题...用栈的思路来做

代码

public class Solution {
    public int evalRPN(String[] tokens) {
        ArrayList<Integer> num  = new ArrayList<Integer>();
        ArrayList<String> ope = new ArrayList<String>();

        for(String oper:tokens)
        {
        	if(oper.equals("+")||oper.equals("-")||oper.equals("*")||oper.equals("/"))
        	{
            	if(oper.equals("+"))
            	{
            		Integer num1 = num.get(num.size()-1);
            		Integer num2 = num.get(num.size()-2);
            		num.remove(num.size()-1);
            		num.remove(num.size()-1);
              		if(num.size()==0)
            			num.add(num1+num2);
              		else
              			num.add(num.size(), num1+num2);
            	}
            	else if(oper.equals("-"))
            	{
            		Integer num1 = num.get(num.size()-1);
            		Integer num2 = num.get(num.size()-2);
            		num.remove(num.size()-1);
            		num.remove(num.size()-1);
              		if(num.size()==0)
            			num.add(num2-num1);
              		else
              			num.add(num.size(), num2-num1);
            	}
            	else if(oper.equals("*"))
            	{
            		Integer num1 = num.get(num.size()-1);
            		Integer num2 = num.get(num.size()-2);
            		num.remove(num.size()-1);
            		num.remove(num.size()-1);
              		if(num.size()==0)
            			num.add( num1*num2);
              		else
              			num.add(num.size(), num1*num2);
            	}
            	else if(oper.equals("/"))
            	{
            		Integer num1 = num.get(num.size()-1);
            		Integer num2 = num.get(num.size()-2);
            		num.remove(num.size()-1);
            		num.remove(num.size()-1);
            		if(num.size()==0)
            			num.add( num2/num1 );
            		else
            			num.add(num.size(), num2/num1);
            		
            	}
        	}
        	else
        	{
        		num.add(Integer.parseInt(oper));
        	}
        }
 
        return num.get(0);
    }
}