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, , 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;
     }
 };