在面试的时候有一道算法题，要求编写一个函数，判断一个给定的字符串s，能否可以由另外两个子串part1和part2合成。 注意，part1和part2中的字母原有顺序必须和s中出现的顺序是一致的。
【示例】
输入：s='hogwarts', part1='hws', part2='ogart'
输出：True
解释：因为hws和ogart的出现顺序和hogwarts一致，合成后也是hogwarts，因此是True
题目难度：困难
题目来源：CodeWars-Merged String Checker
def solution(s: str, part1: str, part2: str)-> bool:
# your code here
assert solution('hogwarts','hws','ogart') is True
assert solution('codewars', 'code', 'wars') is True
assert solution('codewars', 'cod', 'wars') is False
def solution(s: str, part1: str, part2: str) -> bool:
# if len(s) != len(part1) + len(part2):
# return False
cursor1 = cursor2 = 0
for word in s:
if cursor1 < len(part1) and word == part1[cursor1]:
cursor1 += 1
elif cursor2 < len(part2) and word == part2[cursor2]:
cursor2 += 1
else:
return False
return True
def solution(s: str, part1: str, part2: str)-> bool:
# your code here
part1_list = list(part1)
part2_list = list(part2)
for i in range(len(s)):
if len(part1_list) != 0 and s[i] == part1_list[0]:
part1_list.pop(0)
elif len(part2_list) != 0 and s[i] == part2_list[0]:
part2_list.pop(0)
else:
return False
if len(part1_list) == 0 and len(part2_list) == 0:
return True
else:
return False
assert solution('hogwarts','hws','ogart') is True
assert solution('codewars', 'code', 'wars') is True
assert solution('codewars', 'cod', 'wars') is False
assert solution('hogwartsworld','hwsrd','ogartwol') is True
assert solution('hogwartsworld','hwswod','ogartrl') is True
def solution(s: str, part1: str, part2: str)-> bool:
if sorted(s) != sorted(part1+part2):
return False
res1,res2= '',''
for letter in s:
if part1 !='' and part1[0] == letter:
res1 += letter
part1 = part1[1:]
if part2 !='' and part2[0] == letter:
res2 += letter
part2 = part2[1:]
if sorted(res1+res2) == sorted(s):
return True
else:
return False
def solution(s: str, part1: str, part2: str)-> bool:
# your code here
if len(s) != len(part1) + len(part2):
return False
for i in s:
if len(part1) > 0 and i == part1[0]:
part1 = part1.replace(i, '', 1)
elif len(part2) > 0 and i == part2[0]:
part2 = part2.replace(i, '', 1)
else:
return False
else:
return True
assert solution('hogwarts','hws','ogart') is True
assert solution('codewars', 'code', 'wars') is True
assert solution('codewars', 'cod', 'wars') is False
assert solution('sssyyysss', 'ssyys', 'syss') is True
assert solution(‘hwhharts’, ‘hhs’, ‘hwart’) is True
你们的答案执行这个会报错
你这个拆分的不对吧，w要在第二个h的前面，可以分成下面这样的
assert solution(“hwhharts”, “hwhs”, “hart”) is True
assert solution(“hwhharts”, “hhs”, “whart”) is True
def solution(s: str, part1: str, part2: str)-> bool:
# your code here
list1 = []
a = s
if len(s) != len(part1) + len(part2):
return False
else:
for i in range(len(part1)):
list1.append(a.find(part1[i]))
a = a.replace(part1[i],"",1)
print(list1)
for j in part1:
s = s.replace(j,"",1)
print(s)
if sorted(list1) == list1 and s == part2:
return True
else:
return False
assert solution('hogwarts','hws','ogart') is True
assert solution('codewars', 'code', 'wars') is True
assert solution('codewars', 'cod', 'wars') is False
def solution(s: str, part1: str, part2: str)-> bool:
# your code here
if len(s) != len(part1 + part2):
return False
for i in s:
if len(part1) > 0 and part1[0] == i:
part1 = part1.replace(part1[0], "")
elif len(part2) > 0 and part2[0] == i:
part2 = part2.replace(part2[0], "")
else:
return False
return True
assert solution('hogwarts', 'hws', 'ogart') is True
assert solution('codewars', 'code', 'wars') is True
assert solution('codewars', 'cod', 'wars') is False大佬们的用例很犀利 
def solution(s: str, part1: str, part2: str) -> bool:
def func(s: str, part1: str, part2: str):
new_str = ""
for char in s:
if char == part1[0]:
new_str += part1[0]
if len(part1) > 1:
part1 = part1[1::]
else:
new_str += part2[0]
if len(part2) > 1:
part2 = part2[1::]
return new_str == s
return func(s, part1, part2) or func(s, part2, part1)
assert solution('hogwarts', 'hws', 'ogart') is True
assert solution('codewars', 'code', 'wars') is True
assert solution('codewars', 'cod', 'wars') is False
assert solution('sssyyysss', 'ssyys', 'syss') is True
assert solution('hwhharts', 'hhs', 'hwart') is True
assert solution('hogwartsworld','hwsrd','ogartwol') is True
assert solution('hogwartsworld','hwswod','ogartrl') is True
你貌似解决了先part2再part1的情况，但是情况更为复杂就不行了，assert solution(‘ababab’, ‘bab’, ‘aba ’) is True
大佬给的这个断言，我前面的代码是能跑通的~
不过我明白大佬的意思 例如，下面我的这组cases，在座各位无人通过 
：
assert solution('hogwarts','hws','ogart') is True
assert solution('abaaabaa', 'abaa', 'aaab') is True
assert solution('codewars', 'cod', 'wars') is False
assert solution('haah', 'ha', 'ha') is False
······
所以我又写了一个新方法，求更犀利的cases：
def solution(s: str, part1: str, part2: str) -> bool:
if len(s) - len(part1) - len(part2) != 0:
return False
def func(s, part1, part2):
if len(s) == len(part1) == len(part2) == 0:
return True
if len(part1) > 0 and s[0] == part1[0]:
return func(s[1::], part1[1::], part2) or func(s[1::], part2, part1[1::])
elif len(part2) > 0 and s[0] == part2[0]:
return func(s[1::], part1, part2[1::]) or func(s[1::], part2[1::], part1)
return False
return func(s, part1, part2) or func(s, part2, part1)
def solution(s: str, part1: str, part2: str) -> bool:
if len(part1)+len(part2) != len(s):return False
s1,s2 = '',''
for i in s:
if i in part1:s1 += i
if i in part2:s2 += i
for i in part1:
if i not in s1:return False
for i in part2:
if i not in s2:return False
return True
因为 ‘haah’ 只有一个子串 ‘ha’
所以 那条断言不符合题意
子串随便加啊，我是把出问题的点专门精简出来了，下面的字符串符合题意了吗？
assert solution('haahheihei', 'haheihei', 'ha') is False
如果s为codewarts,part1为code，part2为wartss，执行结果就会不对；
另外s有重复字符，比如hologwarts part1为howts,part2为olgar,执行结果也不对
第一个一个错误取消前两行的注释,第二个是本来你输入的就不符合
x, y = 0, 0
for i in range(len(s)):
if x < len(part1) and s[i] == part1[x]:
x = x + 1
continue
elif y < len(part2) and s[i] == part2[y]:
y = y + 1
continue
else:
return False
if len(s) == x + y and x + y == len(part2) + len(part1):
return True
else:
return False
assert solution('haah', 'ha', 'ha') is False
这条用例到底怎么不符合题意了？题目中哪里提到？烦请赐教 @1727375395_3146
另外你说的：
‘haah’ 只有一个子串 ‘ha’
是认真的吗？ 'haah’的子串有：‘h’, ‘a’, ‘ha’, ‘aa’, ‘ah’, ‘haa’, ‘aah’, ‘haah’, ‘’ 呢
最开始用遍历后pop的方法，发现有许多情况没考虑到，最后借鉴大佬们的思路一起套娃。
def solution(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 solution(s[1:], part1[1:], part2) or solution(s[1:], part1, part2[1:])
elif len(part1) > 0 and s[0] == part1[0]:
return solution(s[1:], part1[1:], part2)
elif len(part2) > 0 and s[0] == part2[0]:
return solution(s[1:], part1, part2[1:])
return False