0150. Evaluate Reverse Polish Notation (M)
Evaluate Reverse Polish Notation (M)
题目
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
题意
计算逆波兰表达式(即后缀表达式)的值。
思路
遇到数字就压栈;遇到符号则从栈中弹出两个数字进行运算(注意后出栈的在前,先出栈的在后),将得到的结果再压入栈中;最后栈中只剩一个数,即所求结果。
代码实现
Java
class Solution {
public int evalRPN(String[] tokens) {
Deque<Integer> stack = new ArrayDeque<>();
for (String token : tokens) {
if (token.equals("+")) {
int y = stack.pop(), x = stack.pop();
stack.push(x + y);
} else if (token.equals("-")) {
int y = stack.pop(), x = stack.pop();
stack.push(x - y);
} else if (token.equals("*")) {
int y = stack.pop(), x = stack.pop();
stack.push(x * y);
} else if (token.equals("/")) {
int y = stack.pop(), x = stack.pop();
stack.push(x / y);
} else {
stack.push(Integer.parseInt(token));
}
}
return stack.pop();
}
}
JavaScript
/**
* @param {string[]} tokens
* @return {number}
*/
var evalRPN = function (tokens) {
const stack = []
for (const token of tokens) {
if (!isNaN(token)) {
stack.push(+token)
} else {
const b = stack.pop()
const a = stack.pop()
if (token === '+') stack.push(a + b)
else if (token === '-') stack.push(a - b)
else if (token === '*') stack.push(a * b)
else stack.push(Math.trunc(a / b))
}
}
return stack.pop()
}