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"].
参考：http://blog.csdn.net/doc_sgl/article/details/13064739
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class
Solution {public: void
wordBreakHelper(string s,unordered_set<string> &dict,set<string> &unmatched, int
mn,int
mx, vector<string> &path, vector<string> &res) { int
i = mx < s.length() ? mx : s.length(); for(; i >= mn ; i--) { string preffixstr = s.substr(0,i); if(dict.find(preffixstr) != dict.end()){ path.push_back(preffixstr); if(preffixstr.size() == s.size()) { string tmp = path[0]; for(int
i=1; i<path.size(); i++) tmp+= " "+path[i]; res.push_back(tmp); } string suffixstr = s.substr(i); if(unmatched.find(suffixstr) == unmatched.end()) { int
oldsz = res.size(); wordBreakHelper(suffixstr,dict,unmatched,mn,mx,path,res); if(res.size() == oldsz) unmatched.insert(suffixstr); } path.pop_back(); } } } vector<string> wordBreak(string s, unordered_set<string> &dict) { // Note: The Solution object is instantiated only once. vector<string> res; if(s.size() < 1 || dict.empty()) return
res; unordered_set<string>::iterator it = dict.begin(); int
maxlen=(*it).length(), minlen=(*it).length(); for(it++; it != dict.end(); it++) if((*it).length() > maxlen) maxlen = (*it).length(); else
if((*it).length() < minlen) minlen = (*it).length(); set<string> unmatched; vector<string> path; wordBreakHelper(s,dict,unmatched,minlen,maxlen,path,res); return
res; }}; |
[LeetCode]Word Break II,布布扣,bubuko.com
原文：http://www.cnblogs.com/Rosanna/p/3597249.html