LeetCode刷题实战150:逆波兰表达式求值
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·
2021-01-14 14:29
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.
题意
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
示例 2:
输入: ["4", "13", "5", "/", "+"]
输出: 6
解释: 该算式转化为常见的中缀算术表达式为:(4 + (13 / 5)) = 6
示例 3:
输入: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
输出: 22
解释:
该算式转化为常见的中缀算术表达式为:
((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
解题
public class Solution {
public int evalRPN(String[] tokens) {
LinkedListstack = new LinkedList<>();
for(String string : tokens){
switch (string){
case "+":
stack.push(stack.pop() + stack.pop());
break;
case "-":
stack.push(- stack.pop() + stack.pop());
break;
case "*":
stack.push(stack.pop() * stack.pop());
break;
case "/":
Integer num1 = stack.pop();
Integer num2 = stack.pop();
stack.push(num2 / num1);
break;
default:
stack.push(Integer.parseInt(string));
}
}
return stack.pop();
}
}