问题
This is description of my problem:
- I was thinking about start from left and add one letter and if it's word then check rest if could be separated to words (call recursion function). If yes then I have result if not then I would add one more letter to first word and so on. But this would take first smaller words and I guess it's needed to start with algorithm which check first longer words and then smaller.
- So I was thinking about start with whole word and then recursively check left part without last letter and then right part. If it wouldn't return true then I moved to left with "cursor" and then check left part without 2 last letters and right part (from 2 last letters) and so on. But I think this would required more time then O(n^2).
So can someone help me with basic algoritm which would work and satisfy time complexity? Thanks
回答1:
The dynamic programming approach could be based on the following recursive formula:
f(n) = true
f(i) = OR { d(s[i]s[i+1]...s[j-1]) AND f(j) | for each j such that i < j <= n }
Start from f(0).
Explanation:
If you got to the end of the string - you're good (base clause).
Otherwise, try to split the word wherever you can from the given start (which is i), and recursively invoke on the suffix of the string.
Complexity is actually O(n^2 * g(n)) where g(n) is the time it takes for dict to check if the word is in it. (Every solution need to depend on it somehow...)
Converting to DP solution:
Fill the table from last to first, according to the logic described in the recursive formula.
回答2:
Simple solution in O(n^2) will be :-
DP(n) = (DP(0) && dict(S[1..n]) || (DP(1) && dict(S[2..n]))..... || (DP(n-1) && dict(S[n..n])
DP(0) = true
DP(n) = Whether there is valid sentence made from S[1...n]
S[i...j] is substring from ith index to jth index
来源：https://stackoverflow.com/questions/22917717/dynamic-programming-algorithm-to-repair-text-where-is-all-punctuation-missing