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 if s2 is a scrambled string of s1.

递归来做，也就是s1分为s11和s12，s2分为s21和s22。

判断isScramble(s11,s21)&&isScramble(s12,s22)或者isScramble(s12,s21)&&isScramble(s11,s22)

base case是字符串相同

另外在进入递归前需要剪枝，判断两个字符串是否包含相同的字母，O(n)复杂度。

//http://www.cnblogs.com/ganganloveu/p/4148000.html//http://www.cplusplus.com/reference/string/string/substr/class Solution {public:    bool isScramble(string s1, string s2) {        if(s1 == s2)            return true;                            vector<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;    }};