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"].

dp完以后得到dp数组，然后根据dp数组构造可能的sentence即可。

class Solution {
public:
    int len=0;
    vector<string> v;
    bool* f;
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        word(s,dict);
        if(!f[s.length()])
            return v;
        build(s,s.length(),"",dict);
        return v;
    }
    
    void build(string s,int index,string cur,unordered_set<string> &dict){
        if(index==0){
            v.push_back(cur.substr(0,cur.length()-1));
            return;
        }
        for(int i=1;i<=len && index-i>=0;i++){
            if(find(s.substr(index-i,i),dict) && f[index-i]){
                build(s,index-i,s.substr(index-i,i)+" "+cur,dict);
            }
        }
    }
    
    bool find(string s,unordered_set<string> &dict){
        unordered_set<string>::iterator it = dict.find(s);
        return it!=dict.end();
    }
    void word(string s, unordered_set<string> &dict) {
        string str;
        for(unordered_set<string>::iterator it = dict.begin();it!=dict.end();it++){
            str = *it;
            len = max((int)str.length(),len);
        }
        f = new bool[s.length()+1];
        f[0] = true;
        for(int i=0;i<s.length();i++){
            for(int j=1;i+1-j>=0 && j<=len;j++){
                f[i+1] = f[i+1-j] && find(s.substr(i+1-j,j),dict);
                if(f[i+1])
                    break;
            }
        }
    }
    
};