Leetcode: Word Break

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 as .




第一種方法：遞迴（逾時）

思路：從s的第一個字母向後匹配，如果i前面的首碼可以匹配，就看s字串i以後的尾碼是否匹配





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;    }

第二種方法：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);    }