首先，这些：
def strHead(s):
"""strHead: String -> String"""
return s[0]
def strTail(s, run):
"""strTail: String -> String"""
return s[:]
他们没有做任何有用的事情；第一个返回项目 0，第二个只返回字符串。查克他们。
然后
def runLenAux(s, c):
if s == "":
return 0
else:
if strHead(s)== c:
return 1 + int(runLenAux(strTail(s), c))
else:
return 0
现在有一个明显的问题。当它们充其量是数字列表时，您将它们视为数字。但是其他时候，您将它们视为字符串。
让我们将它们视为列表，因为这样更明显。
所以：
def runLenAux(s, c):
s和是什么c？不知道。我想这是为了获得下一个值，所以做一个[1, 0] → [1, 1, 1, 0]映射。所以让它列出一个清单：
def look_and_say(term):
"""Looks at and says a term."""
->在 Python 3 中，您可以对函数进行注释，而不是做有趣的事情。通过向他们展示这个半隐晦的事实来打动您的同龄人！
def look_and_say(term: "[int]") -> "[int]":
"""Looks at and says a term."""
在这里，我"[int]"用来表示“a listof ints”。
所以，进入下一部分：
if s == "":
return 0
天哪！你打算使用递归吗？！这太多了！我们不是这里的Haskell！
好的，从顶部开始。我们想要遍历，记录我们见过多少相同的项目，是吗？所以我们可以做一个像这样的循环：
for item in term:
count how many are the same
if different, break
（这听起来像是以前做过的事情。嗯，它已经完成了。它被称为itertools.groupby。不过，让我们假装你不知道。）
count = 0
match = None
for item in term:
# If this isn't the same we should stop counting
if match != item:
do_something_with_count_and_match
# Start again from this item
count = 1
match = item
# If this is the same we should continue counting
else:
count = count + 1
这看起来不错！
放什么do_something_with_count_and_match？好吧，让我们保留一份清单，把东西放进去，然后把我们的结果放在那里
# This is what we're building
next_term = []
...
for item in term:
...
if match != item:
# Say "One One" or "Two Threes" or "Three Ones", etc.
next_term.append(count)
next_term.append(match)
# Start again from this item
count = 1
match = item
...
那么，它运行了吗？
def look_and_say(term: "[int]") -> "[int]":
"""Looks at and says a term."""
# This is what we're building
next_term = []
count = 0
match = None
for item in term:
# If this isn't the same we should stop counting
if match != item:
# Say "One One" or "Two Threes" or "Three Ones", etc.
next_term.append(count)
next_term.append(match)
# Start again from this item
count = 1
match = item
# If this is the same we should continue counting
else:
count = count + 1
look_and_say([0])
#>>>
嗯……没有输出。
啊，我们忘记return了next_term：
def look_and_say(term: "[int]") -> "[int]":
... # All that stuff
return next_term
look_and_say([0])
#>>> [0, None]
嗯..那不好。我们需要确保我们没有计算None占位符match：
def look_and_say(term: "[int]") -> "[int]":
...
for item in term:
...
if match != item:
if match is not None:
next_term.append(count)
next_term.append(match)
...
...
我们还需要添加最后一部分，即使它不会触发else（循环将停止）：
def look_and_say(term: "[int]") -> "[int]":
...
for item in term:
...
if match != item:
if match is not None:
next_term.append(count)
next_term.append(match)
return next_term
那就试试吧：
def look_and_say(term: "[int]") -> "[int]":
"""Looks at and says a term."""
# This is what we're building
next_term = []
count = 0
match = None
for item in term:
# If this isn't the same we should stop counting
if match != item:
# Say "One One" or "Two Threes" or "Three Ones", etc.
if match is not None:
next_term.append(count)
next_term.append(match)
# Start again from this item
count = 1
match = item
# If this is the same we should continue counting
else:
count = count + 1
if match is not None:
next_term.append(count)
next_term.append(match)
return next_term
look_and_say([0])
#>>> [1, 0]
look_and_say([1, 0])
#>>> [1, 1, 1, 0]
look_and_say([1, 1, 1, 0])
#>>> [3, 1, 1, 0]
是的！
我这样做是为了向您展示编程不是魔术。你只需要不断尝试和应用你所知道的。
我个人会这样实现它：
from itertools import groupby
# What we just implemented
def look_and_say(string):
for k, v in groupby(string):
yield sum(1 for _ in v)
yield k
list(look_and_say([0]))
#>>> [1, 0]
list(look_and_say([1, 0]))
#>>> [1, 1, 1, 0]
list(look_and_say([1, 1, 1, 0]))
#>>> [3, 1, 1, 0]
但这只是因为我知道groupby。