leetcode 87:扰乱字符串

需要仔细理解题才行，并不是使用二分法的方式进行递归，而是要把字符串分成非空的，也就是一个长abcde的字符串有四种分法

bool scram(std::string s1,std::string s2){
    if(s1.size()!=s2.size())
        return false;
    if(s1==s2)
        return true;
    std::string str1 = s1, str2 = s2;
    sort(str1.begin(), str1.end());
    sort(str2.begin(), str2.end());
    if(str1!=str2)
        return false;
    for(int i=1;i<s1.size();i++){
        std::string m1=s1.substr(0,i);
        std::string m2=s1.substr(i);
        std::string n1 = s2.substr(0, i);
        std::string n2 = s2.substr(i);
        if (scram(m1, n1) && scram(m2, n2))
            return true;
        n1=s2.substr(0,s2.size()-i);
        n2 = s2.substr(s2.size()-i);
        if (scram(m1, n2) && scram(m2, n1))
            return true;
    }
    return false;
}

bool isScramble(std::string s1, std::string s2) {
    if(s1.size()!=s2.size())
        return false;
    return scram(s1,s2);
}