[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;

    }



};

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