leetcode@ [139/140] word break & Word break II

https://leetcode.com/problems/word-break/

Given A string s and a Dictionary of words dict, determine if s can segmented into a space-s Eparated 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" .

classSolution { Public:    BOOLWordbreak (stringS, unordered_set<string>&worddict) {        if(s.length () = =0)return false; Vector<BOOL> Canbreak (s.length (),false);  for(intI=0; I<s.length (); + +i) {if(Worddict.find (S.substr (0, i+1)) !=Worddict.end ()) {Canbreak[i]=true; Continue; }             for(intPre=0;p re<i;++pre) {                if(Canbreak[pre] && worddict.find (S.SUBSTR (pre+1, i-pre))! =Worddict.end ()) {Canbreak[i]=true;  Break; }            }        }                returnCanbreak[s.length ()-1]; }};

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https://leetcode.com/problems/word-break-ii/

Given A string s and a Dictionary of words dict, add spaces in s to construct a sentence where Each of the word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog" ,
dict = ["cat", "cats", "and", "sand", "dog"] .

A solution is ["cats and dog", "cat sand dog"] .

classSolution { Public:    voidFindbreakpoint (vector<BOOL>& Canbreak,string& S, unordered_set<string>&worddict) {         for(intI=0; I<s.length (); + +i) {if(Worddict.find (S.substr (0, i+1)) !=Worddict.end ()) {Canbreak[i]=true; Continue; }             for(intPre=0;p re<i;++pre) {                if(Canbreak[pre] && worddict.find (S.SUBSTR (pre+1, i-pre))! =Worddict.end ()) {Canbreak[i]=true;  Break; }            }        }    }    voidDFS (vector<string>& Res, vector<string>& load, vector<BOOL>& Canbreak,string& S, unordered_set<string>& Worddict,intidx) {if(idx = = S.length ()-1) {             stringTMP ="";  for(intI=0; I<load.size ()-1; ++i) tmp + = Load[i] +" "; if(Load.size () >0) Tmp+=load[load.size ()-1];             Res.push_back (TMP); return; }                 for(intnx=idx+1; Nx<s.length (); + +, X) {            if(Canbreak[nx] && worddict.find (S.SUBSTR (idx+1, NX-IDX))! =Worddict.end ()) {Load.push_back (s.substr (IDX+1, nx-idx));                DFS (res, load, canbreak, S, Worddict, NX);            Load.pop_back (); }}} vector<string> Wordbreak (stringS, unordered_set<string>&worddict) {Vector<BOOL> Canbreak (s.length (),false); Vector<string>res; res.clear (); Vector<string>load; load.clear ();        Findbreakpoint (Canbreak, S, worddict); if(Canbreak[s.length ()-1]) DFS (res, load, canbreak, S, Worddict,-1); returnRes; }};

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[email protected] [139/140] Word Break & Word break II