题解 | #【模板】循环队列#
【模板】循环队列
https://www.nowcoder.com/practice/0a3a216e50004d8bb5da43ad38bcfcbf
思路：给定一个长度为（n+1）的列表，分别以sta_point和end_point作为循环列表的起始指针和终点指针，通过判断两个指针的相对位置来判断是否列表已满或者空。（PS：列表中留有一个不存数字的位置以方便判断）
total_message = input().split(' ')
n = int(total_message[0])
q = int(total_message[1])
lis = [None]*(n+1)
sta_point = 0
end_point = 0
for i in range(q):
opera = input().split(' ')
if len(opera) > 1: # push
# 先判断是否队列满
if (sta_point - 1 == end_point) or (sta_point == 0 and end_point == n):
print('full')
else:
lis[end_point] = opera[1]
end_point = 0 if (end_point == n) else (end_point + 1)
continue
if sta_point == end_point:
print('empty')
continue
if opera[0] == 'front':
print(lis[sta_point])
elif opera[0] == 'pop':
print(lis[sta_point])
sta_point = 0 if (sta_point == n) else (sta_point + 1)
