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【LeetCode】Evaluate Reverse Polish Notation

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
解答

思路:利用栈来存储,遇到数字直接进栈,遇到符号取栈顶的两个数字做运算,最后的结果就是栈顶元素。

public class Solution {
    public int evalRPN(String[] tokens) {
        if(tokens==null||tokens.length==0) return 0;
        Stack<Integer> s=new Stack<Integer>();
        int num1,num2;
        for(int i=0;i<tokens.length;i++){
            String ch=tokens[i];
            if(!(ch.equals("+")||ch.equals("-")||ch.equals("*")||ch.equals("/"))){  //注意是||,不是&&
                s.push(Integer.valueOf(ch));
            }else{
                num1=s.pop();
                num2=s.pop();
                if(ch.equals("+")){
                    s.push(num2+num1);
                }else if(ch.equals("-")){
                    s.push(num2-num1);
                }else if(ch.equals("*")){
                    s.push(num2*num1);
                }else{
                    s.push(num2/num1);
                }
            }
        }
        return s.pop();
    }
}

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