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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.
Some examples:
[“2”, “1”, “+”, “3”, ""] -> ((2 + 1) * 3) -> 9
[“4”, “13”, “5”, “/”, “+”] -> (4 + (13 / 5)) -> 6
解析:
典型的使用栈的例子,本科数据结构课程有讲。
1,遍历列表,建立空栈
2,遇到数字,将它入栈
3,遇到运算符,取栈内数字两次,得到的两个数进行对应运算符运算,将运算结果入栈
4,遍历完毕,取栈,得到最终结果
代码:
def prnval(a):
stack = []
for i in range(len(a)):
if a[i] == '+' or a[i] == '-' or a[i] == '*' or a[i] == '/':
x = stack.pop()
y = stack.pop()
if a[i] == '+':
stack.append(x+y)
elif a[i] == '-':
stack.append(y-x)
elif a[i] == '*':
stack.append(x*y)
else:
stack.append(y//x)
elif a[i] != None:
stack.append(int(a[i]))
return stack.pop()
全部代码:
def prnval(a):
stack = []
for i in range(len(a)):
if a[i] == '+' or a[i] == '-' or a[i] == '*' or a[i] == '/':
x = stack.pop()
y = stack.pop()
if a[i] == '+':
stack.append(x+y)
elif a[i] == '-':
stack.append(y-x)
elif a[i] == '*':
stack.append(x*y)
else:
stack.append(y//x)
elif a[i] != None:
stack.append(int(a[i]))
return stack.pop()
if __name__=='__main__':
a = ["2","1","+","3","*"]
b = ["4", "13", "5", "/", "+"]
print(prnval(a),prnval(b)