[LeetCode] Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given

s = "leetcode",

dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

Solution:

LTE Code:

string srcStr;
unordered_set<string> dic;

bool word_break(int start)
{
    //cout << "Start with " << start << endl;
    if(start >= srcStr.length()) 
        return true;
    for(int i = start;i < srcStr.length();i++)
    {
        string curStr = srcStr.substr(start, i - start + 1);
        if(dic.find(curStr) != dic.end())
        {
            //return word_break(i+1); //The reason is that if current recursion return false, doest not mean the 
            //break should fail, we need to check other combinations.
            if(word_break(i + 1)) 
                return true;
            else 
                continue;
        }
        else continue;
    }

    return false;
}

bool wordBreak(string s, unordered_set<string> &dict) {
        srcStr = s;
        dic = dict;
        return word_break(0);
    }

AC Code:

string srcStr;
unordered_set<string> dic;
int visited[1000][1000];
int checked[1000];

string srcStr;
unordered_set<string> dic;
int visited[1000][1000];
int checked[1000];

bool word_break(int start)
{
    //cout << "Start with " << start << endl;
    if(start >= srcStr.length()) 
        return true;
    for(int i = start;i < srcStr.length();i++)
    {
        string curStr = srcStr.substr(start, i - start + 1);

        int finded = 0;
        if(visited[start][i] == 1)
            finded = true;
        else if(visited[start][i] == 2)
            finded = false;
        else
        {
            finded = (dic.find(curStr) != dic.end()) ? 1 : 2;
            visited[start][i] = finded;
        }

        if(finded == 1)
        {
            if(checked[i + 1] == 1)
                return true;
            else if(checked[i + 1] == 2)
                continue;
            else 
            {
                checked[i + 1] = word_break(i + 1) ? 1 : 2;
                if(checked[i + 1] == 1)
                    return true;
                else
                    continue;
            }
        }
        else continue;
    }

    return false;
}

bool wordBreak(string s, unordered_set<string> &dict) {
        srcStr = s;
        dic = dict;
        for(int i = 0;i < 1000;i++)
            memset(visited[i], 0, sizeof(int) * 1000);
        memset(checked, 0, sizeof(int) * 1000);

        if(s.length() > 1000) return false;

        return word_break(0);
    }