題目：
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
代碼：
class Solution {
public:
bool backScramble(string s1, string s2, map<string, bool>& dp) {
if (s1 == s2)
return true;
int len1 = s1.length(), len2 = s2.length();
if (len1 != len2)
return false;
vector<int> count(26, 0);
for (int i = 0; i < len1; i++)
{
count[s1[i] - 'a']++;
count[s2[i] - 'a']--;
}
for (int i = 0; i < 26; i++)
{
if (count[i] != 0)
return false;
}
bool res = false;
for (int i = 1; i < len1 && !res; i++)
{
res = res || (backScramble(s1.substr(0, i), s2.substr(0, i), dp) && backScramble(s1.substr(i), s2.substr(i), dp)) ||
(backScramble(s1.substr(0, i), s2.substr(len1 - i), dp) && backScramble(s1.substr(i), s2.substr(0, len1 - i), dp));
}
return res;
}
bool isScramble(string s1, string s2) {
map<string, bool> dp;
return backScramble(s1, s2, dp);
}
};