第十五周

87. 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

判断两个字符串是否是Scramble String，及是否是打乱的同一字符串。条件为1.长度相等。2.含有相同的字符，且数量相同。

可以使用递归处理，不断将字符串分割为相同长度的两半再判断是否满足Scramble String，注意：属于头尾不同的两部分如果满足ScrambleString条件的话也可以判定原字符串满足Scramble String。

    bool isScramble(string s1, string s2) {        if(s1==s2)            return true;        if(s1.size()!=s2.size())            return false;        int count[26]={0};        for(int i=0;i<s1.size();i++)        {             count[s1[i]-'a']++;             count[s2[i]-'a']--;        }         for(int i = 0; i < 26; i ++)        {            if(count[i] != 0)                return false;        }         for(int i = 1; i < s1.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(s1.size()-i)) && isScramble(s1.substr(i), s2.substr(0, s1.size()-i)))              )                return true;        }        return false;    }