Scramble String

原题：
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.

解题：
假设字符串只包含小写字母，如果包含其他字符的话也是一样的道理。假设字符串的长度是len,那么可以构建树，但是第一次怎么把单词分成左右子树呢？那么len长度的字符串可以“切”len-1刀，每一“刀”的位置不一样，但都是一分为二，只要有一种满足目标字符串，就可以认为目标字符串是由源字符串变换而来的。递归处理这个过程比较简单。可以AC的C++代码如下：

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