Description:
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Note:
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
题意:计算一个逆波兰式的结果
解法:这道题可以利用栈来求解,遍历字符串,如果是数字则压入栈中,否则,在栈中弹出两个数,进行运算(+、-、*、/)后将结果压入栈中;当遍历完所有的字符串后,栈中的那个元素就是最后所要求解的结果;因为,这里保证了所给的逆波兰式是合法的,所以我们不需要去判断合法性;同时,需要注意的一点是,进行计算的时候,栈中弹出的两个数中首先弹出的是第二操作数,后弹出的是第一操作数;
class Solution {
public int evalRPN(String[] tokens) {
LinkedList stack = new LinkedList<>();
for (String s : tokens) {
if (Character.isDigit(s.charAt(0)) ||
(s.length() > 1 && Character.isDigit(s.charAt(1)))) {
stack.push(s);
} else {
int num2 = Integer.parseInt(stack.pop());
int num1 = Integer.parseInt(stack.pop());
int result = 0;
switch(s.charAt(0)) {
case '+':
result = num1 + num2;
break;
case '-':
result = num1 - num2;
break;
case '/':
result = num1 / num2;
break;
case '*':
result = num1 * num2;
break;
default:
break;
}
stack.push(String.valueOf(result));
}
}
return Integer.parseInt(stack.pop());
}
}