import java.util.Stack;
/**
* 逆波兰表达式求值
* 整数除法只保留整数部分。
* 给定逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
* 示例 1：
* 输入: ["2", "1", "+", "3", "*"]
* 输出: 9
* 解释: 该算式转化为常见的中缀算术表达式为：((2 + 1) * 3) = 9
* 来源：力扣（LeetCode）
* 链接：https://leetcode-cn.com/problems/evaluate-reverse-polish-notation
*/
public class ReversePolishNotation {
public static int evlRPN(String[] tokens){
Stack<Integer> stack = new Stack<>();
for(int i = 0; i < tokens.length; i ++){
if("+".equals(tokens[i])){
stack.push(stack.pop() + stack.pop());
}else if("-".equals(tokens[i])){
Integer num1 = stack.pop();
Integer num2 = stack.pop();
stack.push(num2 - num1);
}else if("*".equals(tokens[i])){
Integer num1 = stack.pop();
Integer num2 = stack.pop();
stack.push(num2 * num1);
}else if("/".equals(tokens[i])){
Integer num1 = stack.pop();
Integer num2 = stack.pop();
stack.push(num2 / num1);
}else {
stack.push(Integer.valueOf(tokens[i]));
}
}
return stack.pop();
}
public static void main(String[] args){
//预期结果 9 6 22
String[] tokens = {"2", "1", "+", "3", "*"};
String[] tokens1 = {"4", "13", "5", "/", "+"};
String[] tokens2 = {"10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"};
System.out.println(evlRPN(tokens));
System.out.println(evlRPN(tokens1));
System.out.println(evlRPN(tokens2));
//实际结果 9 6 22
}
}
Processed:
0.015, SQL:
9
