Leetcode--Word break dynamic planning, detailed understanding

Given A string s and a Dictionary of words dict, determine if s can segmented into a spa Ce-separated sequence of one or more dictionary words.

For example, given

s = "Leetcode",

dict  =  ["leet", "code"] .

Return True because "Leetcode" can be segmented as "Leet code".

Ideas

The first thing I did with that question was to match as much as possible, for example s = "ABCDEFABC" dict=["abc", "Def", and I just took out all the words in the dictionary in S, Finally, if s does not have any letters then it can break;

But the problem comes, s= "aaaaaaa" dict=["AAA", "AAAA", this time will be directly using AAA to divide s into Aaa,aaa,a, thus returning false.

For example, s= "ABCDEEFG" dict=["AB", "CDE", "ee", "CD", "FG", when used in the dictionary "CDE" to split, the result is AB, CDE, E, FG; Thus returning false.

"Dynamic Programming Problem solving"

"Key ★"

Starting with S[2]=c, we've found two dictionary words to match, namely: CDE,CD, we mark the lengths they can stitch together.

ab CDEEFG

AB

Cde

CD

---> Next, we'll start with the location of EFG or EEFG.

"Code"

1  Public classSolution {2      Public BooleanWordbreak (String s, set<string>dict) {3         Boolean[] t =New Boolean[S.length () +1];4T[0] =true;//set first to is true, why?5         //Because We need initial state6  7          for(inti=0; I<s.length (); i++){8             //should continue from match position9             if(!T[i])Ten                 Continue; One   A              for(String a:dict) { -                 intLen =a.length (); -                 intend = i +Len; the                 if(End >s.length ()) -                     Continue; -   -                 if(T[end])Continue; +   -                 if(S.substring (i, End). Equals (a)) { +T[end] =true; A                 } at             } -         } -   -         returnt[s.length ()]; -     } -}

Leetcode--Word break dynamic planning, detailed understanding