140. Word Break II[hard]

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"].

先对单词表建trie树，然后dfs就好

但直接dfs会导致超时。

优化方式是：建一个map，存下搜索过的string s的所有答案，虽然空间消耗大，但是减少了dfs次数

struct node{    vector<node*> a;    bool word;    node()    {        word = false;        a.resize(26,NULL);    }};class Solution {public:    node* root;    set<string> SET;    vector<string> ans;    map<string, vector<string>> mp;        void insert(string word)    {        node* now = root;        for(int i=0;i<word.size();i++)            if(now->a[ word[i]-'a' ]!=NULL)                now = now->a[ word[i]-'a' ];            else            {                now->a[ word[i]-'a' ] = new node();                now = now->a[ word[i]-'a' ];            }        now->word = true;    }        vector<string> getans(string s, node* r)    {    if(mp.find(s)!=mp.end())return mp[s];    vector<string> res;    if(!s.length())    {    res.push_back("");    mp[s] = res;    return mp[s];    }    node* now = r;    int l = s.length();    for(int i = 0 ; i < l ; ++i )    {    if(now->a[ s[i]-'a' ])    now = now->a[ s[i]-'a' ];    else    {    mp[s] = res;    return mp[s];    }    if(now->word)    {    vector<string> nxt = getans(s.substr(i+1),r);    for(auto iter: nxt)    {    string tmp = s.substr(0,i+1) + (iter.length()?" ":"") + iter;    res.push_back(tmp);    }    }    }    mp[s] = res;    return mp[s];    }        vector<string> wordBreak(string s, unordered_set<string>& wordDict){root = new node();        for(auto iter = wordDict.begin();iter!=wordDict.end();iter++)         insert(*iter);        vector<string> ans = getans(s,root);        return ans;    }    };