Evaluate Reverse Polish Notation

最新推荐文章于 2022-03-25 16:52:08 发布

原创最新推荐文章于 2022-03-25 16:52:08 发布 · 403 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 刷题报告专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种使用栈来解决逆波兰表达式求值问题的方法。通过遍历输入的字符串数组，当遇到操作符时从栈顶弹出两个操作数进行计算并将结果压回栈中，最终栈内剩余的元素即为表达式的值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

The key to this problem is to use a stack to store those operands. Once we come with a operator, we use the first two operands on the stack to get a result and put it back to the stack. When there is only one number in the stack, it would be the result. We also have to remember to check whether a token is a number or not.

public class Solution {
    public int evalRPN(String[] tokens) {
        // test whether tokens are empty
        if(tokens.length == 0)
        {
        	System.out.println("No input");
        	return -1;
        }
        // use stack to collect operands
        Stack<Integer> collector = new Stack<Integer>();
        int l = tokens.length;
        for(int i = 0; i < l; i++)
        {
        	if ((tokens[i].equals("+")) || (tokens[i].equals("-")) || (tokens[i].equals("*")) || (tokens[i].equals("/")))
        	{
        	    String str = tokens[i];
        		int operand2 = collector.pop();
        		int operand1 = collector.pop();
        		if(str.equals("+"))
        		{
        		    collector.push(operand1 + operand2);
        		}
        		if(str.equals("-"))
        		{
        		    collector.push(operand1 - operand2);
        		}
        		if(str.equals("*"))
        		{
        		    collector.push(operand1 * operand2);
        		}
        		if(str.equals("/"))
        		{
        		    collector.push(operand1 / operand2);
        		}
        	}
        	else
        	{
        		// we have to test whether this token is a integer or not
        		boolean flag = false;
        		int num = 0;
        		try
        		{
        			num = Integer.parseInt(tokens[i]);
        		}
        		catch(Exception e)
        		{
        			flag = true;
        		}
        		if(flag)
        		{
        			continue;
        		}
        		else
        		{
        			collector.push(num);
        		}
        	}
        }
        // output
        if(collector.size() == 1)
        {
        	return collector.pop();
        }
        return -1;  

    }
}