leetcode140 Word Break II

思路：

直接爆搜会超时，需要使用记忆化搜索。使用map把已经计算过的情况记录下来，避免重复计算。

实现：

class Solution 
{
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) 
    {
         unordered_map<int, vector<string>> dp;
        return dfs(s, 0, wordDict, dp);
    }
    vector<string> dfs(string s, int cur, vector<string>& wordDict, unordered_map<int, vector<string>>& dp)
    {
        if (cur >= s.length())
        {
            vector<string> v;
            v.push_back("");
            return v;
        }
        if (dp.count(cur)) return dp[cur];
        vector<string> ans;
        for (auto it: wordDict)
        {
            int l = it.length();
            if (cur + l <= s.length() && s.substr(cur, l) == it)
            {
                if (cur + l == s.length())
                    ans.push_back(it);
                else
                {
                    vector<string> tmp = dfs(s, cur + l, wordDict, dp);
                    for (auto it1: tmp)
                    {
                        ans.push_back(it + " " + it1);
                    }
                }    
            }
        }
        return dp[cur] = ans;
    }
};