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Copy path150.逆波兰表达式求值.cs
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150.逆波兰表达式求值.cs
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/*
* @lc app=leetcode.cn id=150 lang=csharp
*
* [150] 逆波兰表达式求值
*
* https://leetcode.cn/problems/evaluate-reverse-polish-notation/description/
*
* algorithms
* Medium (53.70%)
* Likes: 545
* Dislikes: 0
* Total Accepted: 189.7K
* Total Submissions: 353.3K
* Testcase Example: '["2","1","+","3","*"]'
*
* 根据 逆波兰表示法,求表达式的值。
*
* 有效的算符包括 +、-、*、/ 。每个运算对象可以是整数,也可以是另一个逆波兰表达式。
*
* 注意 两个整数之间的除法只保留整数部分。
*
* 可以保证给定的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:tokens = ["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:tokens = ["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:tokens = ["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
*
*
*
* 提示:
*
*
* 1 <= tokens.length <= 10^4
* tokens[i] 是一个算符("+"、"-"、"*" 或 "/"),或是在范围 [-200, 200] 内的一个整数
*
*
*
*
* 逆波兰表达式:
*
* 逆波兰表达式是一种后缀表达式,所谓后缀就是指算符写在后面。
*
*
* 平常使用的算式则是一种中缀表达式,如 ( 1 + 2 ) * ( 3 + 4 ) 。
* 该算式的逆波兰表达式写法为 ( ( 1 2 + ) ( 3 4 + ) * ) 。
*
*
* 逆波兰表达式主要有以下两个优点:
*
*
* 去掉括号后表达式无歧义,上式即便写成 1 2 + 3 4 + * 也可以依据次序计算出正确结果。
* 适合用栈操作运算:遇到数字则入栈;遇到算符则取出栈顶两个数字进行计算,并将结果压入栈中
*
*
*/
// @lc code=start
public class Solution {
// 73 57; 85 99 80 39;
public int EvalRPN(string[] tokens) {
int num;
var stk = new Stack<int>();
foreach(string str in tokens){
if(int.TryParse(str, out num)){
stk.Push(num);
}else{
int num1 = stk.Pop();
int num2 = stk.Pop();
switch (str)
{
case "+":
stk.Push(num1 + num2);
break;
case "-":
stk.Push(num2 - num1);
break;
case "*":
stk.Push(num1 * num2);
break;
case "/":
stk.Push(num2 / num1);
break;
default:
break;
}
}
}
return stk.Pop();
}
}
// @lc code=end
/*
["2","1","+","3","*"]\n
["4","13","5","/","+"]\n
["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
*/