Leetcode Word Break II

Given a non-empty string s and a dictionarywordDict containing a list of non-empty words, add spaces ins to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

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"].

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

本题使用DFS进行求解，但不进优化下的DFS无法通过。通过记录中间结果，来避免重复计算，从而更加高效，也就是memoried DFS。

代码如下：

class Solution {public:    unordered_map<string,vector<string>> m;        vector<string> combine(string s,vector<string> pre)    {        for(int i=0;i<pre.size();i++)            pre[i] += " " + s;        return pre;    }        vector<string> wordBreak(string s, vector<string>& wordDict) {        if(m.count(s))            return m[s];        vector<string> result;        if(find(wordDict.begin(),wordDict.end(),s) != wordDict.end())            result.push_back(s);                for(int i=1;i<s.size();i++)        {            string word = s.substr(i);            if(find(wordDict.begin(),wordDict.end(),word) != wordDict.end())            {                string rem = s.substr(0,i);                vector<string> pre = combine(word,wordBreak(rem,wordDict));                result.insert(result.end(),pre.begin(),pre.end());            }        }        m[s] = result;        return result;    }};

运用动态规划进行求解：