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。
说明:
整数除法只保留整数部分。
给定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 <= 104
tokens[i] 要么是一个算符("+"、"-"、"*" 或 "/"),要么是一个在范围 [-200, 200] 内的整数
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。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/evaluate-reverse-polish-notation
python
# 0150.逆波兰表达式
class Solution:
def evalRPN(self, tokens: [str]) -> int:
"""
栈,数字入栈,操作符运算后入栈,最后返回栈顶元素
注意:
- 正负数的判断,正数直接isdigit,负数应除符号外isdigit
- 每次入栈的应该是数字
- 除法运算应特别注意,考虑python2/3除法的特殊性, py3 /根据除的结果是float or int
:param tokens:
:return:
"""
numStack = []
for i in tokens:
if i.isdigit() or i[1:].isdigit():
numStack.append(int(i))
else:
if i == '+':
p1 = numStack.pop()
p2 = numStack.pop()
numStack.append(p2 + p1)
elif i == '-':
p1 = numStack.pop()
p2 = numStack.pop()
numStack.append(p2 - p1)
elif i == '*':
p1 = numStack.pop()
p2 = numStack.pop()
numStack.append(p2 * p1)
elif i == '/':
p1 = numStack.pop()
p2 = numStack.pop()
numStack.append(int(p2 / float(p1)))
return numStack.pop()
if __name__ == "__main__":
tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
tokens1 = ["2","1","+","3","*"]
test = Solution()
print(test.evalRPN(tokens))
print(test.evalRPN(tokens1))
golang
package main
import "strconv"
func evalRPN(tokens []string) int {
var numStack = []int{}
for _, token := range tokens {
val, err := strconv.Atoi(token)
if err == nil {
numStack = append(numStack, val)
} else {
num1, num2 := numStack[len(numStack)-2], numStack[len(numStack)-1]
numStack = numStack[:len(numStack)-2]
switch token {
case "+":
numStack = append(numStack, num1+num2)
case "-":
numStack = append(numStack, num1-num2)
case "*":
numStack = append(numStack, num1*num2)
case "/":
numStack = append(numStack, num1/num2)
}
}
}
return numStack[0]
}