140. Word Break II (String; DP，DFS)

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：利用I的dp判断是否有解，如果有解，使用DFS存储结果（类似 Palindrome Partitioning）

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
        //dp[i]: s[0...i] can be egemented in dict
        //dp[i] = dp[0][k] && d[k][i]
        int len = s.length();
        vector<bool> dp(len,false);
        for(int i = 0; i < len; i++){
            if(wordDict.find(s.substr(0,i+1))!=wordDict.end()) dp[i]=true;
            for(int j = 1; j <= i; j++){
                if((wordDict.find(s.substr(j,i-j+1))!=wordDict.end()) && dp[j-1]) dp[i]=true;
            }
        }
        if(dp[len-1]) dfs(s,0,"",wordDict);
        return ret;
    }
    void dfs(string s, int depth, string strCur, unordered_set<string>& wordDict){
            for(int i = depth; i < s.length(); i++){
                if(wordDict.find(s.substr(depth,i-depth+1))==wordDict.end()) continue;
                if(i==s.length()-1) ret.push_back(strCur+s.substr(depth,i-depth+1));
                else dfs(s,i+1,strCur+s.substr(depth,i-depth+1)+" ", wordDict);
            }
        }
private: 
    vector<string> ret;
};