"Leetcode" Scramble String

Scramble String

Given A string S1, we may represent it as a binary tree by partitioning it to the Non-empty substrings Recursivel Y.

Below is one possible representation of S1 = "great" :

    Great   /      gr    eat/\    /  g   r  E   at           /           a   t

To scramble the string, we are choose any non-leaf node and swap it to the children.

For example, if we choose the node "gr" and swaps its-children, it produces a scrambled string "rgeat" .

    Rgeat   /      RG    eat/\    /  r   G  e   at           /           a   t

We say is "rgeat" a scrambled string of "great" .

Similarly, if we continue to swap the children of nodes "eat" "at" and, it produces a scrambled string "rgtae" .

    Rgtae   /      RG    tae/\    /  r   G  ta  e       /       t   a

We say is "rgtae" a scrambled string of "great" .

Given strings S1 and S2 of the same length, determine if S2 is a scrambled string of S1.

Handed back to do, that is S1 divided into S11 and s12,s2 divided into S21 and S22.

Judge Isscramble (S11,S21) &&isscramble (S12,S22) or isscramble (S12,S21) &&isscramble (S11,S22)

Base case is the same as String

In addition, pruning is required before entering recursion to determine whether two strings contain the same letter, O (n) complexity. Reference http://blog.csdn.net/doc_sgl/article/details/12401335

classSolution { Public:    BOOLIsscramble (stringS1,stringS2) {        //Base Case        if(S1 = =S2)return true; //to here, S1! = S2//Check permutationvector<int> Dict ( -,0);//a~z                 for(inti =0; I < s1.size (); i + +) Dict[s1[i]-'a'] ++;  for(inti =0; I < s2.size (); i + +) Dict[s2[i]-'a'] --;  for(inti =0; I < dict.size (); i + +)            if(Dict[i]! =0)                return false; //to here, S1 and S2 must has same size        intSize =s1.size (); //recursion         for(inti =1; i < size; i + +)        {            if((Isscramble (S1.substr (0, i), S2.substr (0, i)) &&isscramble (S1.substr (i), s2.substr (i)))|| (Isscramble (S1.substr (0, i), S2.substr (size-i)) &&isscramble (S1.substr (i), S2.substr (0, size-i) ))return true; }        return false; }};

"Leetcode" Scramble String