Word Break II

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"].

Analyse:

1. Recursion. Add one character at one time, continuously judge whether the remaining part can be found in the dictionary. If could, then add to the temporary string, if the size of temporary string is bigger than the original string, then we say the string can be segmented by the dictionary items. 

Runtime: TIME LIMIT EXCEEDED. 

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
        vector<string> result;
        if(s.empty() || s.length() == 0) return result;
        
        helper(s, wordDict, 0, "", result);
        return result;
    }
    void helper(string s, unordered_set<string>& wordDict, int start, string temp, vector<string> result){
        if(start >= s.length()){//finish one test
            result.push_back(temp);
            return ;
        }
        string help;
        for(int i = start; i < s.length(); i++){
            help += s[i]; //append one word at a time
            if(wordDict.find(help) != wordDict.end()){
                temp = (temp.length() == 0 ? help : temp + " " + help);
                helper(s, wordDict, i + 1, temp, result);
            }
        }
    }
};

2. Refer to this page. 

　 Runtime: 8ms.

class Solution {
public:
    //2014-2-19 update
    vector<string> wordBreak(string s, unordered_set<string> &dict) 
    {
        vector<string> rs;
        string tmp;
        vector<vector<int> > tbl = genTable(s, dict);
        word(rs, tmp, s, tbl, dict);
        return rs;
    }
    void word(vector<string> &rs, string &tmp, string &s, vector<vector<int> > &tbl,
        unordered_set<string> &dict, int start=0)
    {
        if (start == s.length())
        {
            rs.push_back(tmp);
            return;
        }
        for (int i = 0; i < tbl[start].size(); i++)
        {
            string t = s.substr(start, tbl[start][i]-start+1);
            if (!tmp.empty()) tmp.push_back(' ');
            tmp.append(t);
            word(rs, tmp, s, tbl, dict, tbl[start][i]+1);
            while (!tmp.empty() && tmp.back() != ' ') tmp.pop_back();//tmp.empty()
            if (!tmp.empty()) tmp.pop_back();
        }
    }
    vector<vector<int> > genTable(string &s, unordered_set<string> &dict)
    {
        int n = s.length();
        vector<vector<int> > tbl(n);
        for (int i = n - 1; i >= 0; i--)
        {
            if(dict.count(s.substr(i))) tbl[i].push_back(n-1);
        }
        for (int i = n - 2; i >= 0; i--)
        {
            if (!tbl[i+1].empty())//if we can break i->n
            {
                for (int j = i, d = 1; j >= 0 ; j--, d++)
                {
                    if (dict.count(s.substr(j, d))) tbl[j].push_back(i);
                }
            }
        }
        return tbl;
    }
};