Python: 每日一题 51,Python交流,技术交流,鱼C论坛

ooxx7788 发表于 2017-5-22 19:42:55

Python: 每日一题 51

本帖最后由 ooxx7788 于 2017-5-23 14:30 编辑

简单了几天了，今天是不是要弄点难的呢？那么我们就来点难的吧！5kyu的题目！（不过我感觉还好！）

一个人获得了前往另一个地方的方向指示。指示是这样的，"NORTH", "SOUTH", "WEST", "EAST".
显然"NORTH", "SOUTH"是相反的，"WEST", "EAST"是相反的。连续的两个不同方向的走是无效的。
那么现在就请去除掉这些相对的方向，找出真正的方向吧。

例子：
dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]) => ["WEST"]
dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH"]) => []

注意：
["NORTH", "WEST", "SOUTH", "EAST"]是不可以约除"NORTH", "SOUTH"的哦。"NORTH" and "WEST", "WEST" and "SOUTH", "SOUTH" and "EAST" 不是直接的反方向哦。所以答案就是他本身"NORTH", "WEST", "SOUTH", "EAST"].

a = ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]
test.assert_equals(dirReduc(a), ['WEST'])
u=["NORTH", "WEST", "SOUTH", "EAST"]
test.assert_equals(dirReduc(u), ["NORTH", "WEST", "SOUTH", "EAST"])

**** Hidden Message *****

新手·ing 发表于 2017-5-22 20:19:29

@jerryxjr1220 @lumber2388779 @冬雪雪冬 @SixPy @ooxx7788 @李金龙
大召唤术！！！

ooxx7788 发表于 2017-5-22 20:24:10

新手·ing 发表于 2017-5-22 20:19
@jerryxjr1220 @lumber2388779 @冬雪雪冬 @SixPy @ooxx7788 @李金龙
大召唤术！！！

你在我的帖子里面召唤我！

李金龙 发表于 2017-5-23 10:13:54

新手·ing 发表于 2017-5-22 20:19
@jerryxjr1220 @lumber2388779 @冬雪雪冬 @SixPy @ooxx7788 @李金龙
大召唤术！！！

{:10_266:} 谢邀我不会

冬雪雪冬 发表于 2017-5-23 12:38:24

def dirReduc(lst):
nsew = [["NORTH", "SOUTH"], ["SOUTH", "NORTH"], ["WEST", "EAST"], ["EAST", "WEST"]]
while True:
   for i in range(len(lst) - 1):
         if lst in nsew:
            lst.pop(i + 1)
            lst.pop(i)
            break
   else:
         return lst

jerryxjr1220 发表于 2017-5-23 12:48:28

本帖最后由 jerryxjr1220 于 2017-5-23 12:49 编辑

def dirReduc(lst):
def one_round(lst):
   new = []
   i = 0
   while i < len(lst) - 2:
         if (lst == 'WEST' and lst == 'EAST') or (lst == 'EAST' and lst == 'WEST') or (lst == 'NORTH' and lst == 'SOUTH') or (lst == 'SOUTH' and lst == 'NORTH'):
            i += 2
         else:
            new.append(lst)
            i += 1
   if len(lst) > 1:
         if (lst[-1] == 'WEST' and lst[-2] == 'EAST') or (lst[-1] == 'EAST' and lst[-2] == 'WEST') or (lst[-1] == 'NORTH' and lst[-2] == 'SOUTH') or (lst[-1] == 'SOUTH' and lst[-2] == 'NORTH'):
            pass
         else:
            new.append(lst[-1])
   elif len(lst) < 1:
         return []
   else:
         new.append(lst[-1])
   return new
new = one_round(lst)
while new != lst:
   lst = new
   new = one_round(lst)
return new

jerryxjr1220 发表于 2017-5-23 13:12:29

冬雪雪冬发表于 2017-5-23 12:38

你这个写法比较简洁，借鉴一下。
不过你是用“减法”，我是用的“加法”。
def dirReduc(lst):
def one_round(lst):
   ref = [["NORTH", "SOUTH"], ["SOUTH", "NORTH"],
            ["WEST", "EAST"], ["EAST", "WEST"]]
   new = []
   i = 0
   while i < len(lst):
         if lst in ref:
            i += 2
         else:
            new.append(lst)
            i += 1
   return new
new = one_round(lst)
while new != lst:
   lst = new
   new = one_round(lst)
return new

SixPy 发表于 2017-5-23 19:43:19

很简单~
>>> def dirReduc(ls):
s=({"NORTH", "SOUTH"}, {"WEST", "EAST"})
rslt=[]
for i in ls:
if len(rslt) and ({rslt[-1],i} in s):
rslt.pop()
else:
rslt.append(i)
return rslt

>>> ls=["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]
>>> dirReduc(ls)
['WEST']
>>> dirReduc(["NORTH", "WEST", "SOUTH", "EAST"])
['NORTH', 'WEST', 'SOUTH', 'EAST']
>>> dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH"])
[]
>>>

ooxx7788 发表于 2017-5-23 20:09:11

SixPy 发表于 2017-5-23 19:43
很简单~

大佬首次参与就给这么6的答案

SixPy 发表于 2017-5-24 10:03:01

ooxx7788 发表于 2017-5-23 20:09
大佬首次参与就给这么6的答案

你和新手ing 是同一个人？

ooxx7788 发表于 2017-5-24 10:10:26

SixPy 发表于 2017-5-24 10:03
你和新手ing 是同一个人？

不是啊~

余欲渔 发表于 2017-5-24 10:32:30

def dirReduc(fx):
sc=['NS','SN','WE','EW']
fx= for i in fx]
l=len(fx)
i=0
while i <len(fx)-1:
   if fx+fx in sc:
         fx=fx[:i]+fx
   else :i+=1
if len(fx)==l:
   return fx
else:return dirReduc(fx)

a = ["NORTH", "WEST", "SOUTH","EAST"]
print(dirReduc(a))

solomonxian 发表于 2017-7-11 17:58:17

问题关键在删除中间一对元素后，左右合并的元素有可能相反吧
那递归处理
def dirReduc(a):
direction = [("NORTH","SOUTH"),("WEST","EAST"),("SOUTH","NORTH"),("EAST","WEST")]
lst, length = a[:],len(a)-1# 避免更改输入；控制递归条件

def re_del(lst,length):
   if length == len(lst):# 退出条件是列表长度递归前后无变化
         return lst
   else:
         length = len(lst)
         j=0
         while j < len(lst)-1:
            if (lst,lst) in direction:
               lst.pop(j)# 删除会影响index
               lst.pop(j)
            else:
               j += 1
         return re_del(lst,length)
return re_del(lst,length)

张大象 发表于 2017-9-28 17:08:59

本帖最后由张大象于 2017-9-28 17:19 编辑

Direct=['NORTH','WEST','EAST','SOUTH']
def dirReduc(L):
for i in range(len(L)//2):
for j in range(len(L)):
if j+1 == len(L):
break
else:
if Direct.index(L)+Direct.index(L) == 3:
L.pop(j+1)
L.pop(j)
break
return L
a = ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]
print(dirReduc(a))

shigure_takimi 发表于 2017-12-4 15:37:38

本帖最后由 shigure_takimi 于 2017-12-4 15:38 编辑

def hasOppositeDir(directions): # 任意两个相邻的相反的返回True
oppositeDir = {'NORTH':'SOUTH', 'EAST':'WEST', 'SOUTH':'NORTH', 'WEST':'EAST'}
length = len(directions)
if length < 2:
   return False
else:
   for index in range(length -1):
         if oppositeDir] == directions:
            return True
   return False

def getRealDir(directions):
oppositeDir = {'NORTH':'SOUTH', 'EAST':'WEST', 'SOUTH':'NORTH', 'WEST':'EAST'}
index = 0
while hasOppositeDir(directions):
   if oppositeDir] == directions:
         directions.pop(index)
         directions.pop(index) # pop掉index的方向后，index+1的方向变为index，pop掉。
   else:
         index += 1
   if index > len(directions)-2:
         index = 0
return directions

a = ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]
b = ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH"]

print(getRealDir(a))
print(getRealDir(b))

##符合预期。
##['WEST']
##[]
##看了别人写的，发现自己的比较笨呢。

tsembrace 发表于 2018-4-26 14:53:58

'''
一个人获得了前往另一个地方的方向指示。指示是这样的，"NORTH", "SOUTH", "WEST", "EAST".
显然"NORTH", "SOUTH"是相反的，"WEST", "EAST"是相反的。连续的两个不同方向的走是无效的。
那么现在就请去除掉这些相对的方向，找出真正的方向吧。
示例：
dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]) => ["WEST"]
dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH"]) => []
'''
def mystrip(ds):
while '--' in ds:
   ds = ds.replace('--','-')

return ds

def dirReduc(dl):
GS = ['SOUTH-NORTH','NORTH-SOUTH','WEST-EAST','EAST-WEST']
ds = '-'.join(dl)
while GS in ds or GS in ds or GS in ds or GS in ds:
   if GS in ds:
         ds = ''.join(ds.split(GS))
         ds = mystrip(ds)
   if GS in ds:
         ds = ''.join(ds.split(GS))
         ds = mystrip(ds)
   if GS in ds:
         ds = ''.join(ds.split(GS))
         ds = mystrip(ds)
   if GS in ds:
         ds = ''.join(ds.split(GS))
         ds = mystrip(ds)
   dl = ds.strip('-').split('-')
return dl

print(dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]))

小强工作室 发表于 2018-7-23 18:38:29

理解了

咕咕鸡鸽鸽 发表于 2019-1-9 19:40:43

kankan

鱼小发表于 2019-1-10 13:54:44

吧

jluzhpanda 发表于 2019-8-24 11:33:11

lst_1 = ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]
lst_2 = ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH"]
lst_3 = ["NORTH", "WEST", "SOUTH", "EAST"]
def way(lst):
a = 0
while a < len(lst):
   lst2 = []
   for i in range(1, len(lst)):
         # print(i)
         if lst == 'NORTH' and lst == 'SOUTH':
            lst, lst = 0, 0
         elif lst == 'SOUTH' and lst == 'NORTH':
            lst, lst = 0, 0
         elif lst == 'WEST' and lst == 'EAST':
            lst, lst = 0, 0
         elif lst == 'EAST' and lst == 'WEST':
            lst, lst = 0, 0
   for i in lst:
         if i != 0:
            lst2.append(i)
   a += 1
   lst = lst2
return (lst)
print (way(lst_1))
print (way(lst_2))
print (way(lst_3))
'''
['WEST']
[]
['NORTH', 'WEST', 'SOUTH', 'EAST']
'''

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Python: 每日一题 51