s 可以由其他两个字符串part1 和part2 组成。
限制是part1 和中的字符应与part2 中的字符顺序相同s 。
魔法师给你下面的例子。
例如：
'hogwarts' 由 'hgr' and 'owats' 合并:
s: h o g w a r t s = hogwarts
part1: h g r = hgr
part2: o w a t s = owats
题目难度：中等
题目来源： https://www.codewars.com/kata/54c9fcad28ec4c6e680011aa
def is_merge(s: str, part1: str, part2: str) -> bool:
"""your code"""
assert is_merge("hogwarts","hgr","owats") == True
assert is_merge("hogwarts","hog","warts") == True
assert is_merge("hogwarts","hgr","wats") == False
def is_merge(s: str, part1: str, part2: str) -> bool:
"""your code"""
str2 = part1+part2
if len(s) == len(str2):
ls =[i for i in s]
for i in str2:
if i in ls:
ls.remove(i)
if len(ls) == 0:
return True
else:
return False
else:
return False
if __name__ == '__main__':
assert is_merge("hogwarts","hgr","owats") == True
assert is_merge("hogwarts","hog","warts") == True
assert is_merge("hogwarts","hgr","wats") == False
assert is_merge("hogwarts", "hgr", "wssaaats") == False
assert is_merge("hogwarts", "hgr", "waoas") == False
joyoko
2022 年6 月 29 日 05:09
3
# daily_0415
def is_merge(s: str, part1: str, part2: str) -> bool:
if len(s) <= 1:
return s == part1 + part2
elif len(part1) > 0 and len(part2) > 0 and s[0] == part1[0] and s[0] == part2[0]:
return is_merge(s[1:], part1[1:], part2) or is_merge(s[1:], part1, part2[1:])
elif len(part1) > 0 and s[0] == part1[0]:
return is_merge(s[1:], part1[1:], part2)
elif len(part2) > 0 and s[0] == part2[0]:
return is_merge(s[1:], part1, part2[1:])
return False
xiaopeng
2022 年6 月 29 日 08:09
4
def is_merge(s: str, part1: str, part2: str) -> bool:
for i,v in enumerate(s):
if len(part1) != 0 and ord(v)==ord(part1[0]):
part1 = part1[1:]
continue
if len(part2) != 0 and ord(v)==ord(part2[0]):
part2 = part2[1:]
continue
return False
return True
assert is_merge("hogwarts","hgr","owats") == True
assert is_merge("hogwarts","hog","warts") == True
assert is_merge("hogwarts","hgr","wats") == False
def is_merge(s: str, part1: str, part2: str) -> bool:
#递归
if not part1:
return s == part2
if not part2:
return s == part1
if not s:
return part1+part2 == ''
if s[0]==part1[0] and is_merge(s[1:],part1[1:],part2):
return True
if s[0]==part2[0] and is_merge(s[1:],part1,part2[1:]):
return True
return False
assert is_merge("hogwarts","hgr","owats") == True
assert is_merge("hogwarts","hog","warts") == True
assert is_merge("hogwarts","hgr","wats") == False
def is_merge(s: str, part1: str, part2: str) -> bool:
"""your code"""
if len(s) != len(part1) + len(part2):
return False
if (len(part1) == 1 and s == part1) or (len(part2) == 1 and s == part2):
return True
if len(part1) > 0 and len(part2) > 0 and s[0] == part1[0] and s[0] == part2[0]:
return is_merge(s[1:], part1[1:], part2) or is_merge(s[1:], part1, part2[1:])
if len(part1) > 0 and s[0] == part1[0]:
return is_merge(s[1:], part1[1:], part2)
if len(part2) > 0 and s[0] == part2[0]:
return is_merge(s[1:], part2[1:], part1)
assert is_merge("hogwarts", "hgr", "owats") == True
assert is_merge("hogwarts", "hog", "warts") == True
assert is_merge("hogwarts", "hgr", "wats") == False
assert is_merge("hhgwaatss", "hgwas", "hats") == True
qunzi
2023 年3 月 22 日 02:35
8
def is_merge(s: str, part1: str, part2: str) -> bool:
s=list(s)
if len(part1+part2)==len(s):
for i in part1:
s.remove(i)
return True if "".join(s)==part2 else False
else:
return False
assert is_merge("hogwarts","hgr","owats") == True
assert is_merge("hogwarts","hog","warts") == True
assert is_merge("hogwarts","hgr","wats") == False
应该有点问题，如果part1中的字符顺序和s中的不一致，也会返回True；
lirish
2023 年3 月 23 日 10:29
11
建议加上长度的判断，比如assert is_merge(“hogwarts”,“hog”,“wartsas”)是不成立的
jordan
2023 年3 月 24 日 06:57
13
def is_merge_2(s: str, part1: str, part2: str) -> bool:
"""solution #2 deal with ordering of part1 and part2 strings"""
if len(part1 + part2) == len(s):
index1, index2 = 0, 0
for i in range(len(s)):
if index1 < len(part1) and s[i] == part1[index1]:
index1 += 1
elif index2 < len(part2) and s[i] == part2[index2]:
index2 += 1
else:
return False
return True
else:
return False
def main():
assert is_merge_2("hogwarts", "hgr", "owats") == True
assert is_merge_2("hogwarts", "hog", "warts") == True
assert is_merge_2("hogwarts", "hgr", "wats") == False
assert is_merge_2("hogwarts", "hgrt", "woas") == False
if __name__ == "__main__":
main()
lekaixin
2023 年12 月 3 日 06:08
14
def is_merge(s: str, part1: str, part2: str) -> bool:
if len(s)!=len(part1)+len(part2):
return False
else:
index_list1=[s.index(part1[i]) if part1[i] in s else 'x' for i in range(len(part1))]
index_list2=[s.index(part2[i]) if part2[i] in s else 'x' for i in range(len(part2))]
if 'x' not in index_list1 and 'x' not in index_list2:
if sorted(index_list1)==index_list1 and sorted(index_list2)==index_list2:
index_res_list=sorted(index_list1+index_list2)
if index_res_list == [i for i in range(len(s))]:
return True
else:
return False
else:
return False
else:
return False
assert is_merge("hogwarts","hgr","owats") == True
assert is_merge("hogwarts","hog","warts") == True
assert is_merge("hogwarts","hgr","wats") == False
assert is_merge("hogwarts","hgo","warts") == False
assert is_merge("hogwarts","hog","awrts") == False
进入魔法世界之前，您要面临一个挑战，编写一个算法来检查给定的字符串 , 是否