题目：
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.
Example 1:
Input: s1 = "great", s2 = "rgeat" Output: true
Example 2:
Input: s1 = "abcde", s2 = "caebd" Output: false
代码：
方法一（用了排序算法的递归)：
class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1.size() != s2.size()) return false;
if (s1 == s2) return true;
string str1 = s1, str2 = s2;
sort(str1.begin(), str1.end());
sort(str2.begin(), str2.end());
if (str1 != str2) return false;
for (int i = 1; i < s1.size(); ++i) {
string s11 = s1.substr(0, i);
string s12 = s1.substr(i);
string s21 = s2.substr(0, i);
string s22 = s2.substr(i);
if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
s21 = s2.substr(s1.size() - i);
s22 = s2.substr(0, s1.size() - i);
if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
}
return false;
}
};方法二(没用排序算法的递归):
class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1 == s2) return true;
if (s1.size() != s2.size()) return false;
int n = s1.size(), m[26] = {0};
for (int i = 0; i < n; ++i) {
++m[s1[i] - 'a'];
--m[s2[i] - 'a'];
}
for (int i = 0; i < 26; ++i) {
if (m[i] != 0) return false;
}
for (int i = 1; i < n; ++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(n - i)) && isScramble(s1.substr(i), s2.substr(0, n - i)))) {
return true;
}
}
return false;
}
};