我仍然很难理解算法中的所有步骤，我应该如何处理输入字符串的所有可能分区，而没有基于字典的空格。
我读到我们必须使用TRIE结构使用一些动态编程和反向跟踪（深度优先搜索），但我从概念上不了解它是如何工作的。
首先我想要的是：
input
string = `theology`
dictionnary
dict = [the , theology]
output i am expecting
[the, o, l, o, g, y ] ,[theology]
关于我“这里的错误理解是我看到的步骤”（我使用R语言，但我更需要解释逻辑而不是特定的语言答案）
0 - Transform my dictionnary into a trie
```
library("triebeard")
trie <- trie(keys = c("the", "theology","toto","what"),
values = c("the", "theology",
"toto","what"))
```
1 - calculate the number of character on the string
- On this case I have 8.
2 - Take character by character n+1 and look inside the trie
2.1 I take the first character is t so i look at t and do the longest match
It return NA so i go to the next character
```
longest_match(trie, "t")
```
2.2 Next char `"th"` returns NA
2.3 Next char "the" returns `"the"` so i add it the a list
3a - Now i don't know what to do should I continue to the 4 position take one character
look at the longest match see NA and so on so forth
3b - Or should go and look at `"theo"`
我猜这与前缀匹配有关，而不是与最长的匹配。
如果您可以指导我如何使用trie获得预期的输出
[更新]
为了更好地理解我的问题，这里有一个可能的解决方案在python使用字典，我发现/修改了感谢interjay的回答：https://stackoverflow.com/a/2174156/5348895
wordList = ['the','theology']
words = set( s.lower() for s in wordList )
def splitString(s):
found = []
def rec(stringLeft, wordsSoFar):
if not stringLeft:
found.append(wordsSoFar)
for pos in xrange(1, len(stringLeft)+1):
if stringLeft[:pos] in words:
rec(stringLeft[pos:], wordsSoFar + [stringLeft[:pos]])
elif (pos==len(stringLeft)) and stringLeft[:pos] not in wordList and len(stringLeft)<len(s):
rec(stringLeft[1:], wordsSoFar + [stringLeft[:1]])
rec(s.lower(), [])
return found
output = splitString('theology')
for x in output:
print ','.join(x)
由于