這道題一眼看去，就死類似於單詞拆分i的深度優先搜尋就可以解決的題目，不過這裡公升級了，就是要把每一種切分結果都要返回。
具體**如下:
class
solution
:def
wordbreak
(self, s:
str, worddict: list[
str])-
> list[
str]
: mem =
tmp = self.dfs(s, worddict, mem)
res =
for item in tmp:
' '.join(item)
)return res
defdfs(self, s, worddict, mem)
:if self.check(s, worddict)
==false
:return
if s in mem.keys():
return mem[s]
if s =='':
return[[
]]res =
for i in
range(1
,len
(s)+1)
:if s[
:i]in worddict and self.check(s[i:
], worddict)
: substr = self.dfs(s[i:
], worddict, mem)
for item in substr:
[s[:i]
]+ item)
return res
defcheck
(self, s, worddict)
:if s in worddict:
return
true
index_set =
[false]*
(len
(s)+1)
index_set[0]
=true
for i in
range(1
,len
(s)+1)
:# i [1, 8]
for idx in
range
(i):
# idx [0, 7]
if index_set[i]
==false
and index_set[idx]
and s[idx:i]
in worddict:
index_set[i]
=true
return index_set[-1
]
2.正向dfs，利用反向判斷的資訊可以快速判斷後面是否可以分解，從而避免不必要的計算量；
具體**如下：
class
solution
:'''反向判斷'''
defwordbreak
(self, s:
str, worddict: list[
str])-
> list[
str]
: judges = self.get_judgement(s, worddict)
if judges[0]
==false
:return
mem =
res =
tmp = self.dfs(s, worddict, mem, judges)
for item in tmp:
' '.join(item)
)return res
defdfs(self, s, worddict, mem, judges)
:if s in mem.keys():
return mem[s]
if s =='':
return[[
]]res =
for i in
range(1
,len
(s)+1)
:#先判斷字典中是不是存在這個字串，然後判斷這個字串後面是否能分解
if s[
:i]in worddict and judges[i]
: substr = self.dfs(s[i:
], worddict, mem, judges[i:])
for item in substr:
[s[:i]
]+ item)
mem[s]
= res
return res
defget_judgement
(self, s, worddict)
:#len(s) = 5
res =
[false]*
(len
(s)+1)
res =
[false]*
(len
(s)+
1)
res[-1
]=true
for i in
range
(len
(s))[:
:-1]
:#i 7 to 0
for ptr in
range
(i+1
,len
(s)+1)
[::-
1]:#ptr 8 to i+1
if res[i]
==false
and res[ptr]
and s[i:ptr]
in worddict:
res[i]
=true
return res
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