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

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.

给出两个字符串，判断第二个字符串是否能从第一个字符串“搅乱“得来。“搅乱“的规则是对每个字符串递归的进行旋转（不旋转也行）两个子字符串。用递归的方法即可。通过递归划分子问题：对当前两个字符串s1，s2，如果s1(0,i)和s2(0,i)符合“搅乱”规则且s1(i,n1)和s2(i,n2)符合“搅乱”规则（交换），或者s1(0,i)和s2(n2-i,n2)符合“搅乱”规则且s1(i,n1)和s2(0,i)符合“搅乱”规则,则说明当前两个字符串符合“搅乱”规则。在递归过程中为了减少递归次数，通过检测两个字符串字符数和所含字符是否全都一样来剪枝。

代码：

class Solution
{
public:
bool isScramble(string s1, string s2)
{
if(s1 == s2) return true;
int n1 = s1.size(), n2 = s2.size();
if(n1 != n2) return false;

int cnt[256] = {0};
for(int i = 0; i < n1; ++i)
{
++cnt[s1[i]];
--cnt[s2[i]];
}
for(int i = 0; i < 256; ++i)
{
if(cnt[i] != 0)
{
return false;
}
}

for(int i = 1; i < n1; ++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(n2 - i)) && isScramble(s1.substr(i), s2.substr(0, n2 - i))) )
{
return true;
}
}
return false;
}
};