Longest Uncommon Subsequence I &

昨天在LeetCode上搞了两题，如下：

Longest Uncommon Subsequence I

没啥好说的，水过去

int findLUSlength(string a, string b) {
    if (a == b) return -1;
    else return a.length()>b.length()? a.length(): b.length();
}

Longest Uncommon Subsequence II

第二题，把第一题省下的时间又搞回去了。。
没啥精妙的套路，暴力求解，有点长。

bool sContainsT(string s, string t) {
    bool isContain = false;
    int i = 0, j = 0;
    while (i < s.length() && j < t.length()) {
        if (t[j] == s[i]) {
            j++; i++;
        }
        else 
            i++;
    }
    if (j == t.length())
        isContain = true;
    return isContain;
}

int findLUSlength(vector<string>& strs) {
    vector<vector<string> > stringByLenght(11);
    set<string> stringSet;
    set<string> uniqueStringSet;
    set<string>::iterator itr;
    int length = 0, currStringLength = 0;
    int longest = strs[0].length();
    for (size_t i = 0; i < strs.size(); i++) {
        stringByLenght[strs[i].length()].push_back(strs[i]);
        if (strs[i].length() > longest)
            longest = strs[i].length();
    }
    for (int i = 0; i < stringByLenght[longest].size(); i++) {
        if (stringSet.insert(stringByLenght[longest][i]).second == true) {
            length += longest;
        }
        else {
            length -= longest;
            if (length < 0) length = 0;
        }
    }
    if (length > 0)
        return longest;
    else
        length = -1;
    currStringLength = longest - 1;
    while (currStringLength > 0 && length == -1) {
        uniqueStringSet.clear();
        for (int i = 0; i < stringByLenght[currStringLength].size(); i++) {
            if (stringSet.insert(stringByLenght[currStringLength][i]).second == true) 
                uniqueStringSet.insert(stringByLenght[currStringLength][i]);
            else 
                uniqueStringSet.erase(stringByLenght[currStringLength][i]);
        }
        if (uniqueStringSet.size() > 0) {
            for (itr = uniqueStringSet.begin(); itr != uniqueStringSet.end(); itr++) {
                for (int i = 0; i < stringByLenght[longest].size(); i++)
                {
                    if (sContainsT(stringByLenght[longest][i], *itr)) {
                        length = -1;
                        break;
                    }
                    else
                        length = currStringLength;
                }
            }
        }
        currStringLength--;
    }
    return length;
}

// test case.
int main()
{
    vector<string> strs;
    int a = 0;
    string strings[] = {"j","j","viez","ogk","ogk","lfn","ypmhwx","ypmhwx","m","m","ak","ivivzoncju","ivivzoncju","wmybi","wmybi","dyzfjg","dyzfjg"};
    //string strings[] = { "a","b","c", "a", "b" };
    //string strings[] = {"aaaaa", "aaaaa", "aaa", "aaa", "aaaaa", "aabaa"};
    //string strings[] = { "aaa", "aaa", "aa"};
    for (int i = 0; i < 17; i++) {
        strs.push_back(strings[i]);
    }
    a = findLUSlength(strs);
    cout << a << endl;
    return 0;
}

3ms AC，时间还可以，主要是限定的字符串长度短，个数少。不过绝对不是一个好方法。