Word Break 2

Given a string s and a dictionary of wordsdict, add spaces ins 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"]

分析：本题可以采用动态规划解决。。

但是同时要求把所有的可能的路径输出。

所以，在动态规划的过程中，要将路径的上一步保存下来，从而便于路径的恢复。

class Solution {public:    vector<string> wordBreak(string s, unordered_set<string> &dict) {        vector<bool> f(s.size()+1, false);        vector<vector<bool> > prev(s.size()+1, vector<bool>(s.size()));        f[0] = true; // empty string                for(int i=1; i<=s.size(); ++i)        {          for(int j=i-1; j>=0; --j){              if(f[j] && dict.find(s.substr(j, i-j)) != dict.end()){                  f[i] = true;                  prev[i][j] = true;              }          }        }        vector<string> path;        vector<string> result;        genPath(s, prev, s.size(), path, result);                return result;    }private:    void genPath(const string& s, const vector<vector<bool> >&prev, int cur, vector<string>&path, vector<string>&result){        if(cur == 0){            string tmp;            for(auto iter = path.rbegin(); iter != path.rend(); ++iter)              tmp += *iter + ' ';            tmp.erase(tmp.end()-1);            result.push_back(tmp);        }        for(int j=0; j<s.size(); ++j){            if(prev[cur][j]){                path.push_back(s.substr(j, cur-j));                genPath(s, prev, j, path, result);                path.pop_back();            }        }    }};