LeetCode---Word Break 2

Given a string s and a dictionary of wordsdict, 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"]

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();
            }
        }
    }
};