leetcode-Evaluate the value of an arithmetic expression in Reverse Polish Notation

Posted 2020-09-25 银河末班车

leetcode 逆波兰式求解

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

import java.util.Stack;
public class Solution {
     public int evalRPN(String[] tokens) {
         Stack<Integer> stack = new Stack<Integer>();
        int tmp;
        for(int i = 0; i<tokens.length; i++){
            if("+".equals(tokens[i])){
                int a = stack.pop();
                int b = stack.pop();
                stack.add(b+a);
            }
            else if("-".equals(tokens[i])){
                int a = stack.pop();
                int b = stack.pop();
                stack.add(b-a);
            }
            else if("*".equals(tokens[i])){
                int a = stack.pop();
                int b = stack.pop();
                stack.add(b*a);
            }
            else if("/".equals(tokens[i])){
                int a = stack.pop();
                int b = stack.pop();
                stack.add(b/a);
            }
            else{
                stack.add(Integer.parseInt(tokens[i]));
            }
            
            
        }
        return stack.pop();
     }
}

根据你波兰式求值。看到逆波兰式可以想到栈，扫描表达式，遇到数字则将数字入栈，遇到运算符，时，则从栈顶弹出两个元素，后弹出的元素在运算符的左边，先弹出的元素在元素符的右边，执行运算，将结果入栈。扫描结束后，栈中的元素只剩下一个，即逆波兰式的值