[Leetcode]Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great   /    \  gr    eat / \    /  \g   r  e   at           / \          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

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

    rgeat   /    \  rg    eat / \    /  \r   g  e   at           / \          a   t

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

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

    rgtae   /    \  rg    tae / \    /  \r   g  ta  e       / \      t   a

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

Given two strings s1 and s2 of the same length, determine ifs2 is a scrambled string of s1.

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class Solution {public:    /*algorithm        native idea: after several times swap, s2 can change to s1        and there is swap point, we just enumerate each swap point         and compare s11,s21 and s12 and s22        this is recrusive procedure.    */    bool isScramble(string s1, string s2) {        if(s1.size() != s2.size())return false;        if(s1==s2)return true;        int tbl[26]={0};        for(int i = 0;i < s1.size();i++){            tbl[s1[i]-'a']++;            tbl[s2[i]-'a']--;        }        for(int k=0;k < 26;k++){            if(tbl[k])return false;        }        for(int l=1;l < s1.size();l++){            string s11=s1.substr(0,l);            string s12=s1.substr(l);            string s21=s2.substr(0,l);            string s22=s2.substr(l);            if(isScramble(s11,s21)&&isScramble(s12,s22))return true;            s21=s2.substr(s2.size()-l);            s22=s2.substr(0,s2.size()-l);            if(isScramble(s11,s21)&&isScramble(s12,s22))return true;        }        return false;    }};