[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.

递归剪枝。

对两个字符串可以判断排序后是否相等确认是否是scrambled的，如果不相等一定返回false,剪枝后速度加快很多。

递归过程，把s1和s2分成两个部分，sa1,sb1,sa2,sb2.

返回 ((sa1~sb1)&&(sa2~sb2))||((sa1~sb2)&&(sa2~sb1)).

(s1~s2)表示是否scrambled

class Solution {
public:
bool isScramble(string s1, string s2) {
int len1 = s1.length();
int len2 = s2.length();
if(len1!=len2)
return false;
if(s1==s2)
return true;
int A[26] = {0};
for(int i=0; i<len1; ++i)
++A[s1[i]-'a'];
for(int i=0; i<len2; ++i)
--A[s2[i]-'a'];
for(int i=0; i<26; ++i){
if(A[i]!=0)
return false;
} //递归剪枝，判断字符串各个字母是否相同，采用类似桶排序简化。
for(int i=1; i<len1; ++i){
bool ret = isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i));
ret = ret||(isScramble(s1.substr(0,i),s2.substr(len2-i,i))&&isScramble(s1.substr(i),s2.substr(0,len2-i)));//递归解
if(ret)
return true;
}
return false;
}
};