[LeetCode] Word Break II

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"].

Solution:

递归过程注意用空间换时间，以免TLE。

class Solution {
public:
    vector<string>* res;
int* visited;
string srcStr;

vector<string> word_break(int start, string s, unordered_set<string> &dict)
{
        if(visited[start] == 1)
            return res[start];

        vector<string> ans;

        int len = s.length();
        if(len == 0) 
        {
            visited[start] = 1;
            res[start] = ans;
            return ans;
        }
        //enumerate the first space location
        for(int i = 1;i <= len;i++)
        {
            //i is the number of char in the first string
            string firstStr = s.substr(0, i);
            string secondStr = s.substr(i, len - i);

            if(dict.find(firstStr) != dict.end())
            {
                //first string can be found, get the break result of the second string
                vector<string> secondRes = word_break(start + i, secondStr, dict);
                
                if(i == len)
                {
                    ans.push_back(firstStr);
                }
                else
                {
                    if(secondRes.size() != 0)
                    {
                        for(int j = 0;j < secondRes.size();j++)
                        {
                            string cur = firstStr + " " + secondRes[j];
                        //    cout << "cur = " <<  cur << endl;
                            ans.push_back(cur);
                        }
                    }
                }
            }
        }    

        res[start] = ans;
        visited[start] = 1;
        return ans;
}

vector<string> wordBreak(string s, unordered_set<string> &dict) {
        srcStr = s;
        vector<string> ans;
        if(s.length() == 0) return ans;
        res = new vector<string>[s.length() + 1];
        visited = new int[s.length() + 1];
        memset(visited, 0, (s.length() + 1) * sizeof(int));

        ans = word_break(0, s, dict);

        return ans;
    }

};