LeetCode刷题MEDIM篇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.

Note:

• Division between two integers should truncate toward zero.
• The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.

Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation: 
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

十分钟尝试

和上一个题目类似，都是用stack维持顺序，遇到运算符出stack进行运算，注意减法和除法的顺序问题。另外这个题目比上一个其实简单，上一个题目记录前一个的运算符，比较别扭。类似的思路，写过了，很不错

class Solution {
    public int evalRPN(String[] tokens) {
        if(tokens.length==0) return 0;
        Deque<Integer> stack=new LinkedList();
        for(int i=0;i<tokens.length;i++){
            String curr=tokens[i];
            if(curr.equals("+")){
                int op1=(Integer)stack.pop();
                int op2=(Integer)stack.pop();
                stack.push(op1+op2);
            }
            //注意顺序，减法和除法
            else if(curr.equals("-")){
                int op1=(Integer)stack.pop();
                int op2=(Integer)stack.pop();
                stack.push(op2-op1);
            }
            else if(curr.equals("*")){
                int op1=(Integer)stack.pop();
                int op2=(Integer)stack.pop();
                stack.push(op1*op2);
            }
            else if(curr.equals("/")){
                int op1=(Integer)stack.pop();
                int op2=(Integer)stack.pop();
                stack.push(op2/op1);
            }
            else{
                stack.push(Integer.valueOf(curr));
            }
      
          }
                         return stack.getLast();
                       }
}