如何实现 process_deque(commands) 来测试我的 Deque 函数？
how to Implement process_deque(commands) to test my Deque function?
首先，我需要实现一个 Deque 函数。
push_front 向双端队列的头部添加一个键
push_back 在双端队列的尾部添加一个键
pop_front 从双端队列的头部提取一个键并return保存它
pop_back 从双端队列的尾部提取一个键并return保存它
front returns head element without removed it
back returns 尾元素而不删除它
size returns 双端队列中元素的个数
clear 删除双端队列中的所有元素
使用此方法和错误处理实现 Deque class。对于 pop_frontpop_front、pop_backpop_back，前、后方法必须检查双端队列中是否有元素。如果它为空，则该方法必须 return 字符串“error”而不是数值。
push_front, push_back, clear方法有到 return 字符串 "ok".
最终实施 process_deque(commands)，它接受命令列表和 returns 每个命令的答案列表。
class Deque:
def __init__(self, max_len = 60000):
self.max_len = max_len + 1
self.queue = [0] * self.max_len
self.head = 0
self.tail = 0
def push_front(self, key):
self.queue[self.head] = key
self.head= (self.head - 1) % self.max_len
return "ok"
def push_back(self, key):
self.queue[self.tail] = key
self.tail = (self.tail + 1) % self.max_len
return "ok"
def pop_front(self):
if self.head == self.tail:
return "error"
else:
res = self.queue[self.head]
self.head = (self.head + 1) % self.max_len
return res
def pop_back(self):
res = self.queue[self.tail]
self.tail = (self.tail - 1) % self.max_len
return res
def front(self):
if self.head == self.tail:
return "error"
else:
return self.queue[self.head]
def back(self):
if self.head == self.tail:
return "error"
else:
return self.queue[self.tail]
def clear(self):
self.queue = []
return "ok"
def size(self):
return len(self.queue)
def process_deque(commands):
if __name__ == "__main__": # ***This is the 3 test case***
test_cmd = ["push_front 1", "push_front 2", "push_back 6", "front", "back", "clear", "size", "back"]
# should print ["ok", "ok", "ok", 2, 6, "ok", 0, "error"]
print(process_deque(test_cmd))
test_cmd = ["pop_front", "back", "push_back 2", "size"]
# should print ["error", "error", "ok", 1]
print(process_deque(test_cmd))
test_cmd = ["push_back 1", "push_front 10", "push_front 4", "push_front 5", "back", "pop_back", "pop_back", "back"]
# should print ["ok", "ok", "ok", "ok", 1, 1, 10, 4]
print(process_deque(test_cmd))
我不确定如何实施 process_deque（命令）来测试案例。
谢谢。
我只是调整了我的双端队列函数，下面的代码更干净，因为我不知道如何实现 process_deque(命令)，所以我根本没有测试它。
class Deque:
def __init__(self):
self.item = []
def push_front(self, key):
self.item.append(key)
return "ok"
def push_back(self, key):
self.items.insert(0,key)
return "ok"
def pop_front(self):
if self.head == self.tail:
return "error"
else:
self.items.pop()
def pop_back(self):
return self.items.pop(0)
def front(self):
if self.head == self.tail:
return "error"
else:
return self.item[0]
def back(self):
if self.head == self.tail:
return "error"
else:
return self.item[-1]
def clear(self):
self.queue = []
return "ok"
def size(self):
return len(self.queue)
您可以使用 getattr 方法获取上述方法 here
def process_deque(commands):
d = Deque()
output = []
for each_command in commands:
splitted = each_command.split(" ")
method_name = splitted[0]
arg = splitted[1] if len(splitted)==2 else None
method = getattr(d, method_name)
if(arg):
output.append(method(arg))
else:
output.append(method())
return output
但是您的 deque class 方法实现 不正确 。
这不在这个问题的范围内，所以为了证明上述方法是正确的，我使用 python list 本身可以充当 deque.
class Deque:
def __init__(self, max_len = 60000):
# self.max_len = max_len + 1
# self.queue = [0] * self.max_len
# self.head = 0
# self.tail = 0
self.queue = []
def push_front(self, key):
# self.queue[self.head] = key
# self.head= (self.head - 1) % self.max_len
self.queue.insert(0, key)
return "ok"
def push_back(self, key):
# self.queue[self.tail] = key
# self.tail = (self.tail + 1) % self.max_len
self.queue.append(key)
return "ok"
def pop_front(self):
# if self.head == self.tail:
# return "error"
# else:
# res = self.queue[self.head]
# self.head = (self.head + 1) % self.max_len
# return res
if len(self.queue) == 0:
return "error"
else:
res = self.queue.pop(0)
return res
def pop_back(self):
# res = self.queue[self.tail]
# self.tail = (self.tail - 1) % self.max_len
if len(self.queue) == 0:
return "error"
else:
res = self.queue.pop()
return res
return res
def front(self):
if len(self.queue) == 0:
return "error"
else:
return self.queue[0]
def back(self):
if len(self.queue) == 0:
return "error"
else:
return self.queue[-1]
def clear(self):
self.queue.clear()
return "ok"
def size(self):
return len(self.queue)
def process_deque(commands):
d = Deque()
output = []
for each_command in commands:
splitted = each_command.split(" ")
method_name = splitted[0]
arg = splitted[1] if len(splitted)==2 else None
method = getattr(d, method_name)
if(arg):
output.append(method(arg))
else:
output.append(method())
return output
if __name__ == "__main__": # ***This is the 3 test case***
test_cmd = ["push_front 1", "push_front 2", "push_back 6", "front", "back", "clear", "size", "back"]
# should print ["ok", "ok", "ok", 2, 6, "ok", 0, "error"]
print(process_deque(test_cmd))
test_cmd = ["pop_front", "back", "push_back 2", "size"]
# should print ["error", "error", "ok", 1]
print(process_deque(test_cmd))
test_cmd = ["push_back 1", "push_front 10", "push_front 4", "push_front 5", "back", "pop_back", "pop_back", "back"]
# should print ["ok", "ok", "ok", "ok", 1, 1, 10, 4]
print(process_deque(test_cmd))
输出
['ok', 'ok', 'ok', '2', '6', 'ok', 0, 'error']
['error', 'error', 'ok', 1]
['ok', 'ok', 'ok', 'ok', '1', '1', '10', '4']
首先，我需要实现一个 Deque 函数。
push_front 向双端队列的头部添加一个键
push_back 在双端队列的尾部添加一个键
pop_front 从双端队列的头部提取一个键并return保存它
pop_back 从双端队列的尾部提取一个键并return保存它
front returns head element without removed it
back returns 尾元素而不删除它
size returns 双端队列中元素的个数
clear 删除双端队列中的所有元素
使用此方法和错误处理实现 Deque class。对于 pop_frontpop_front、pop_backpop_back，前、后方法必须检查双端队列中是否有元素。如果它为空，则该方法必须 return 字符串“error”而不是数值。
push_front, push_back, clear方法有到 return 字符串 "ok".
最终实施 process_deque(commands)，它接受命令列表和 returns 每个命令的答案列表。
class Deque:
def __init__(self, max_len = 60000):
self.max_len = max_len + 1
self.queue = [0] * self.max_len
self.head = 0
self.tail = 0
def push_front(self, key):
self.queue[self.head] = key
self.head= (self.head - 1) % self.max_len
return "ok"
def push_back(self, key):
self.queue[self.tail] = key
self.tail = (self.tail + 1) % self.max_len
return "ok"
def pop_front(self):
if self.head == self.tail:
return "error"
else:
res = self.queue[self.head]
self.head = (self.head + 1) % self.max_len
return res
def pop_back(self):
res = self.queue[self.tail]
self.tail = (self.tail - 1) % self.max_len
return res
def front(self):
if self.head == self.tail:
return "error"
else:
return self.queue[self.head]
def back(self):
if self.head == self.tail:
return "error"
else:
return self.queue[self.tail]
def clear(self):
self.queue = []
return "ok"
def size(self):
return len(self.queue)
def process_deque(commands):
if __name__ == "__main__": # ***This is the 3 test case***
test_cmd = ["push_front 1", "push_front 2", "push_back 6", "front", "back", "clear", "size", "back"]
# should print ["ok", "ok", "ok", 2, 6, "ok", 0, "error"]
print(process_deque(test_cmd))
test_cmd = ["pop_front", "back", "push_back 2", "size"]
# should print ["error", "error", "ok", 1]
print(process_deque(test_cmd))
test_cmd = ["push_back 1", "push_front 10", "push_front 4", "push_front 5", "back", "pop_back", "pop_back", "back"]
# should print ["ok", "ok", "ok", "ok", 1, 1, 10, 4]
print(process_deque(test_cmd))
我不确定如何实施 process_deque（命令）来测试案例。
谢谢。
我只是调整了我的双端队列函数，下面的代码更干净，因为我不知道如何实现 process_deque(命令)，所以我根本没有测试它。
class Deque:
def __init__(self):
self.item = []
def push_front(self, key):
self.item.append(key)
return "ok"
def push_back(self, key):
self.items.insert(0,key)
return "ok"
def pop_front(self):
if self.head == self.tail:
return "error"
else:
self.items.pop()
def pop_back(self):
return self.items.pop(0)
def front(self):
if self.head == self.tail:
return "error"
else:
return self.item[0]
def back(self):
if self.head == self.tail:
return "error"
else:
return self.item[-1]
def clear(self):
self.queue = []
return "ok"
def size(self):
return len(self.queue)
您可以使用 getattr 方法获取上述方法 here
def process_deque(commands):
d = Deque()
output = []
for each_command in commands:
splitted = each_command.split(" ")
method_name = splitted[0]
arg = splitted[1] if len(splitted)==2 else None
method = getattr(d, method_name)
if(arg):
output.append(method(arg))
else:
output.append(method())
return output
但是您的 deque class 方法实现 不正确 。
这不在这个问题的范围内，所以为了证明上述方法是正确的，我使用 python list 本身可以充当 deque.
class Deque:
def __init__(self, max_len = 60000):
# self.max_len = max_len + 1
# self.queue = [0] * self.max_len
# self.head = 0
# self.tail = 0
self.queue = []
def push_front(self, key):
# self.queue[self.head] = key
# self.head= (self.head - 1) % self.max_len
self.queue.insert(0, key)
return "ok"
def push_back(self, key):
# self.queue[self.tail] = key
# self.tail = (self.tail + 1) % self.max_len
self.queue.append(key)
return "ok"
def pop_front(self):
# if self.head == self.tail:
# return "error"
# else:
# res = self.queue[self.head]
# self.head = (self.head + 1) % self.max_len
# return res
if len(self.queue) == 0:
return "error"
else:
res = self.queue.pop(0)
return res
def pop_back(self):
# res = self.queue[self.tail]
# self.tail = (self.tail - 1) % self.max_len
if len(self.queue) == 0:
return "error"
else:
res = self.queue.pop()
return res
return res
def front(self):
if len(self.queue) == 0:
return "error"
else:
return self.queue[0]
def back(self):
if len(self.queue) == 0:
return "error"
else:
return self.queue[-1]
def clear(self):
self.queue.clear()
return "ok"
def size(self):
return len(self.queue)
def process_deque(commands):
d = Deque()
output = []
for each_command in commands:
splitted = each_command.split(" ")
method_name = splitted[0]
arg = splitted[1] if len(splitted)==2 else None
method = getattr(d, method_name)
if(arg):
output.append(method(arg))
else:
output.append(method())
return output
if __name__ == "__main__": # ***This is the 3 test case***
test_cmd = ["push_front 1", "push_front 2", "push_back 6", "front", "back", "clear", "size", "back"]
# should print ["ok", "ok", "ok", 2, 6, "ok", 0, "error"]
print(process_deque(test_cmd))
test_cmd = ["pop_front", "back", "push_back 2", "size"]
# should print ["error", "error", "ok", 1]
print(process_deque(test_cmd))
test_cmd = ["push_back 1", "push_front 10", "push_front 4", "push_front 5", "back", "pop_back", "pop_back", "back"]
# should print ["ok", "ok", "ok", "ok", 1, 1, 10, 4]
print(process_deque(test_cmd))
输出
['ok', 'ok', 'ok', '2', '6', 'ok', 0, 'error']
['error', 'error', 'ok', 1]
['ok', 'ok', 'ok', 'ok', '1', '1', '10', '4']