共计 1320 个字符，预计需要花费 4 分钟才能阅读完成。
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
难度：medium
题目：求算术表达式在反波兰表示法中的值。有效的操作符是 +、-、*、/。每个操作数可以是一个整数或另一个表达式。
思路：栈
Runtime: 6 ms, faster than 91.00% of Java online submissions for Evaluate Reverse Polish Notation.Memory Usage: 36.6 MB, less than 100.00% of Java online submissions for Evaluate Reverse Polish Notation.
class Solution {
public int evalRPN(String[] tokens) {
Stack<Integer> stack = new Stack<>();
int tNum = 0;
for (int i = 0; i < tokens.length; i++) {
switch(tokens[i]) {
case “+” :
stack.push(stack.pop() + stack.pop());
break;
case “-” :
tNum = stack.pop();
stack.push(stack.pop() – tNum);
break;
case “*” :
stack.push(stack.pop() * stack.pop());
break;
case “/” :
tNum = stack.pop();
stack.push(stack.pop() / tNum);
break;
default :
stack.push(Integer.parseInt(tokens[i]));
}
}
return stack.pop();
}
}
