[LeetCode] 150. Evaluate Reverse Polish Notation Java

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

题意及分析:给出一个字符串,程序计算逆波兰式的结果,遍历表达式,碰到操作数入栈,碰到操作符就从栈顶取出两个操作数,再将计算后的结果入栈,最后栈中剩余的唯一操作数就是计算结果。 这里注意的是string类型判断值是否相等需要用.equals。

代码:

public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> stack=new Stack<>();
        for(int i=0;i<tokens.length;i++){
        	String token=tokens[i];
        	if(token.equals("+")||token.equals("-")||token.equals("*")||token.equals("/")){
        		int op1=stack.pop();
        		int op2=stack.pop();
        		if(token.equals("+")){
        			stack.push(op2+op1);
        		}
        		if(token.equals("-")){
        			stack.push(op2-op1);
        		}
        		if(token.equals("*")){
        			stack.push(op2*op1);
        		}
        		if(token.equals("/")){
        			stack.push(op2/op1);
        		}
        	}else{
        		stack.push(Integer.parseInt(token));
        	}
        }
        int res=stack.peek();
        // System.out.println(res);
        return res;
    }
}

  

原文地址:https://www.cnblogs.com/271934Liao/p/7026407.html