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"].
思路：
思路类似于Word Break，不过在那题的基础上增加了一个vector<vector<string> > result, 用来记录从 i 出发的字符串可能表示的种类。
代码：
1 vector<vector<string> > result; 2 vector<int> canForm; 3 void wordBreak(string s, unordered_set<string> &dict, int start, int min, int max){ 4 if(canForm[start] != -1){ 5 return; 6 } 7 int len = s.length(); 8 if(len < start+min){ 9 canForm[start] = 0; 10 return; 11 } 12 int i, j; 13 for(i = min; i <= max; i++){ 14 if(start + i == len){ 15 break; 16 } 17 string tmp = s.substr(start, i); 18 if(dict.find(tmp) != dict.end()){ 19 if(canForm[start+i] == -1) 20 wordBreak(s, dict, start+i, min, max); 21 if(canForm[start+i] == 1){ 22 canForm[start] = 1; 23 for(j = 0; j < result[start+i].size(); j++){ 24 result[start].push_back(tmp+" "+result[start+i][j]); 25 } 26 } 27 } 28 } 29 if(start + i == len){ 30 string tmp = s.substr(start); 31 if(dict.find(tmp) != dict.end()){ 32 canForm[start] = 1; 33 result[start].push_back(tmp); 34 } 35 } 36 if(canForm[start] == -1) 37 canForm[start] = 0; 38 } 39 vector<string> wordBreak(string s, unordered_set<string> &dict) { 40 // IMPORTANT: Please reset any member data you declared, as 41 // the same Solution instance will be reused for each test case. 42 int len = s.length(); 43 result = vector<vector<string> >(len); 44 canForm = vector<int>(len, -1); 45 int max = INT_MIN; 46 int min = INT_MAX; 47 for(unordered_set<string>::iterator i = dict.begin(); i != dict.end() ; i++){ 48 int length = i->length(); 49 max = max > length ? max : length; 50 min = min < length ? min : length; 51 } 52 if(len < min) 53 return result[0]; 54 wordBreak(s, dict, 0, min, max); 55 return result[0]; 56 }
第二遍深搜：
1 bool search(string s, int index, int canBreak[], unordered_set<string> &dict, vector<string> &res, string tmp){ 2 if(index == s.length()){ 3 res.push_back(tmp); 4 return true; 5 } 6 if(canBreak[index] == 0) 7 return false; 8 for(int i = index; i < s.length(); i++){ 9 string t = s.substr(index, i-index+1); 10 if(dict.find(t) != dict.end()){ 11 string cur = tmp; 12 if(tmp == "") 13 tmp = t; 14 else 15 tmp = tmp + " " + t; 16 if(search(s, i+1, canBreak, dict, res, tmp)) 17 canBreak[index] = 1; 18 tmp = cur; 19 } 20 } 21 if(canBreak[index] == -1) 22 canBreak[index] = 0; 23 return canBreak[index] == 1; 24 } 25 vector<string> wordBreak(string s, unordered_set<string> &dict) { 26 // IMPORTANT: Please reset any member data you declared, as 27 // the same Solution instance will be reused for each test case. 28 vector<string> res; 29 string tmp = ""; 30 int l = s.length(); 31 if(l == 0) 32 return res; 33 int canBreak[l+1]; 34 memset(canBreak, -1, sizeof(canBreak)); 35 canBreak[l] = 1; 36 search(s, 0, canBreak, dict, res, tmp); 37 return res; 38 }