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 if s2 is a scrambled string of s1.

有一种方法叫做简单粗暴。。。

递归+剪枝

bool isScramble(string s1, string s2) {
// Note: The Solution object is instantiated only once.
if(s1.length() != s2.length()) return false;
if(s1 == s2) return true;

int A[26] = {0};
for(int i = 0; i < s1.length(); i++)
++A[s1[i]-'a'];

for(int j = 0; j < s2.length(); j++)
--A[s2[j]-'a'];

for(int k = 0; k < 26; k++)
if(A[k] != 0) return false;

for(int i = 1; i < s1.length(); i++)
{
bool result = isScramble(s1.substr(0,i), s2.substr(0,i))
&& isScramble(s1.substr(i), s2.substr(i));
result = result || (isScramble(s1.substr(0,i), s2.substr(s2.length()-i, i))
&& isScramble(s1.substr(i), s2.substr(0,s2.length()-i)));
if(result) return true;
}
return false;
}