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 Subscribe to see which companies asked this question.
题意：求逆波兰表达式的结果
public class Solution {
public int isopnd(String s){
if(s.length()!=1)return 0;
char c=s.charAt(0);
switch(c){
case '+': return 1;
case '-': return 2;
case '*': return 3;
case '/': return 4;
default:return 0;
}
}
public int evalRPN(String[] tokens) {
if(tokens.length==1) return Integer.valueOf(tokens[0]);
Stack<Integer> stack=new Stack<Integer>();
int ans=0;
for(String tk : tokens){
int op=isopnd(tk);
if(op!=0){
int a=stack.peek();
stack.pop();
int b=stack.peek();
stack.pop();
if(op==1){
ans=a+b;
}
if(op==2){
ans=b-a;
}
if(op==3){
ans=a*b;
}
if(op==4){
ans=b/a;
}
System.out.println(ans);
stack.push(ans);
}else{
stack.push(Integer.valueOf(tk));
}
}
return ans;
}
}PS：用栈。第一次调用栈。遇到数字，入栈，遇到运算符，出栈俩数字做计算，结果再入栈。
