Word break & word break II

Word break

Given a stringSAnd a dictionary of wordsDict, Determine ifSCan be segmented into a space-separated sequence of one or more dictionary words.

For example, givenS="leetcode",Dict=["leet", "code"].

Return true because"leetcode"Can be segmented"leet code".

Note: For deep search, you must remember the results of each visit (the screen is recorded here ).

bool judge(string s, unordered_set<string> &dict, vector<bool> &tag) {    if(s == "") return true;    for(int i = 1; i <= s.length(); ++i) {        if(tag[s.size()-i] && dict.find(s.substr(0, i)) != dict.end())  {            if (judge(s.substr(i, s.size()-i), dict, tag)) return true;            else tag[s.size()-i] = 0;         }    }    return false;}class Solution {public:    bool wordBreak(string s, unordered_set<string> &dict) {        if(s == "") return true;        vector<bool> tag(s.size()+1, true); //the value is the result that (index) length of reserved string can return;        return judge(s, dict, tag);    }};

 

Word break II

Given a stringSAnd a dictionary of wordsDict, Add spaces inSTo construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, givenS="catsanddog",Dict=["cat", "cats", "and", "sand", "dog"].

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

Note: The method is clever. Remember all the ending positions starting from each position that can become the input string. Then, perform a deep search.

void dfs(string s, vector<vector<int> > & Reach, int Id, string path, vector<string> &vec) {if(Id == s.size()) { vec.push_back(path); return; }for(size_t i = 0; i < Reach[Id].size(); ++i) {path = path + (Id == 0 ? s.substr(Id, Reach[Id][i]) : " " + s.substr(Id, Reach[Id][i]-Id));dfs(s, Reach, Reach[Id][i], path, vec);path.erase(path.end()-(Id == 0 ? Reach[Id][i] : (Reach[Id][i]-Id+1)), path.end()); }}class Solution {public:vector<string> wordBreak(string s, unordered_set<string> &dict) {vector<string> vec;int n = s.size();if(n == 0) return vec;vector<vector<int> > reachable(n, vector<int>());for(int end = n; end > 0; --end) {if(end < n && reachable[end].empty()) continue;for(int start = 0; start < end; ++start) {if(dict.find(s.substr(start, end-start)) != dict.end()) reachable[start].push_back(end);}}dfs(s, reachable, 0, string(""), vec);return vec;}};

 

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