140. Word Break II

缓存的重要性：
方法一：直接dfs，TLE

class Solution {
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        vector<string> res;
        helper(s, 0, dict, "", res);
        return res;
    }
private:
    void helper(string& s, int idx, unordered_set<string>& dict, string temp, vector<string>& res) {
        if (idx == s.size()) {
            res.push_back(temp.substr(0, temp.size()-1));
            return;
        }
        for (int i = idx; i < s.size(); ++i) {
            string str = s.substr(idx, i-idx+1);
            if (dict.find(str) != dict.end())
                helper(s, i+1, dict, temp+str+" ", res);
        }
    }
};

方法二：缓存，16ms

class Solution {
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        unordered_map<int, vector<string>> m;
        auto res = helper(s, 0, dict, m);
        for (auto& str : res) 
            str = str.substr(0, str.size()-1);
        return res;
    }
    
private:
    vector<string> helper(string& s, int idx, unordered_set<string>& dict, unordered_map<int, vector<string>>& m) {
        //m存的是s中某个位置向后的所能够组成的string的集合
        if (idx == s.size())
            return vector<string>{string("")};//最后这个必须要有，不然最后一个结果都没有
        if (m.find(idx) != m.end())
            return m[idx];
        vector<string> res;
        for (int i = idx; i < s.size(); ++i) {
            string str = s.substr(idx, i-idx+1);
            if (dict.find(str) != dict.end()) {
                auto nextStrs = helper(s, i+1, dict, m);
                for (auto& nextStr : nextStrs)
                    res.push_back(str + " " + nextStr);
            }
        }
        m[idx] = res;
        return res;
    }
};
//必须要缓存中间结果，不然超时
//可以缓存的不一定是dp

所以缓存不是一定得在dp上用，其他为了提高效率也可以用