Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
,
.
Can be segmented.
Method 1: recursion (timeout)
Idea: the first letter of s is matched backward. If the prefix before I can be matched, it depends on whether the suffix after string I matches
bool wordBreak(string s, unordered_set<string> &dict) { // Note: The Solution object is instantiated only once. if(s.length() < 1) return true;bool flag = false;for(int i = 1; i <= s.length(); i++){string tmpstr = s.substr(0,i);unordered_set<string>::iterator it = dict.find(tmpstr);if(it != dict.end()){if(tmpstr.length() == s.length())return true;flag = wordBreak(s.substr(i),dict);}if(flag)return true;}return false; }
Method 2: dp
bool wordBreakHelper(string s, unordered_set<string> &dict,set<string> &unmatch) { if(s.length() < 1) return true;bool flag = false;for(int i = 1; i <= s.length(); i++){string prefixstr = s.substr(0,i);unordered_set<string>::iterator it = dict.find(prefixstr);if(it != dict.end()){string suffixstr = s.substr(i);set<string>::iterator its = unmatch.find(suffixstr);if(its != unmatch.end())continue;else{flag = wordBreakHelper(suffixstr,dict,unmatch);if(flag) return true;else unmatch.insert(suffixstr);}}}return false; }bool wordBreak(string s, unordered_set<string> &dict) { // Note: The Solution object is instantiated only once. int len = s.length();if(len < 1) return true;set<string> unmatch;return wordBreakHelper(s,dict,unmatch); }