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.
第二次写这个题了，上次为了不反复生成string对象，用下标来做的，确实要复杂一些。
http://blog.csdn.net/a83610312/article/details/8858378
这次干脆图简单在递归中反复生成string对象，从提交的结果来看，还不是很慢，但其实还是建议用下标来做。
class Solution {
public:
bool isScramble(string s1, string s2) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (s1.compare(s2)==0)
return true;
string t1(s1),t2(s2);
sort(t1.begin(),t1.end());
sort(t2.begin(),t2.end());
if ( t1.compare(t2)!=0)
return false;
int n=s1.length();
for(int i=1;i<n;i++)
{
string ls1=s1.substr(0,i);
string rs1=s1.substr(i,n-i);
string ls2=s2.substr(0,i);
string rs2=s2.substr(i,n-i);
if ( isScramble(ls1,ls2)&&isScramble(rs1,rs2))
return true;
string ls22=s2.substr(0,n-i);
string rs22=s2.substr(n-i,i);
if ( isScramble(ls1,rs22)&&isScramble(rs1,ls22))
return true;
}
return false;
}
};