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Word Break II
Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each 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"]
.
class Solution { public: vector<string> wordBreak(string s, unordered_set<string> &dict) { vector<string> result; if(check(s,dict)==false) return result; int v[s.length()+1];v[0]=0; generate(result,s,dict,v,1); return result; } bool check(string s, unordered_set<string> &dict) { bool able[s.length()+1]; able[0]=true; for(int i=1;i<=s.length();i++) { able[i]=false; for(int j=0;j<i;j++) if(able[j] && dict.find(s.substr(j,i-j))!=dict.end()) { able[i]=true; continue; } } return able[s.length()]; } void generate(vector<string>& result,const string& s,const unordered_set<string> &dict,int* v,int vdep) { if(v[vdep-1]==s.length()) { string snew=s.substr(0,v[1]); for(int i=2;i<vdep;i++) { snew=snew+" "+s.substr(v[i-1],v[i]-v[i-1]); } result.push_back(snew); return; } for(int i=1;i<=s.length()-v[vdep-1];i++) if(dict.find(s.substr(v[vdep-1],i))!=dict.end()) { v[vdep]=v[vdep-1]+i; generate(result,s,dict,v,vdep+1); } } };
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